HZNU第十二届校赛赛后补题
愉快的校赛翻皮水！
题解
A 温暖的签到，注意用gets
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x); #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x); #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 1000000 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; string a; int main(){ while(getline(cin,a)){ a.back() = '!'; cout << a << endl; } return 0; }
B.比赛的时候一直以为是主席树上操作或者其他的高级数据结构,万万没想到是在序列特性下手,打一张最小的不冲突的表就会发现斐波那契数列是最小不冲突序列,int范围内最多容纳47个数左右,所以小于50的范围暴力查询,大于50的范围必定YES
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x); #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x); #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 2e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; LL a[maxn]; LL b[maxn]; int main(){ Sca2(N,M); for(int i = 1; i <= N ; i ++) Scl(a[i]); for(int i = 1; i <= M ; i ++){ int l,r; Sca2(l,r); if(r - l + 1 < 3){ puts("NO"); continue; } if(r - l + 1 >= 50) puts("YES"); else{ int flag = 0; for(int j = l; j <= r; j ++) b[j] = a[j]; sort(b + l,b + r + 1); for(int j = l + 2; j <= r; j ++){ if(b[j - 2] + b[j - 1] > b[j]){ flag = 1; break; } } if(flag) puts("YES"); else puts("NO"); } } return 0; //1 1 2 3 5 8 13 21 34 }
D. 8说了,温暖的签到
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x); #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x); #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 2e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; int a[maxn]; set<int>Q; int main(){ N = read(); while(N--) Q.insert(read()); Pri(Q.size()); return 0; //1 1 2 3 5 8 13 21 34 }
E.给一个条件构造的图，求图上的哈密顿回路。
可以猜想到N为奇数的时候始终不可行。
N为偶数的时候必定可行，并且可以发现，每个点都恰好有两个出度和两个入度，
这就可以转换成欧拉回路直接做
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x); #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x); #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 10010; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; int ans[maxn],cnt; bool vis[maxn]; void dfs(int t){ int l = (t * 2) % N,r = (t * 2 + 1) % N; if(l > r) swap(l,r); if(!vis[r]){vis[r] = 1;dfs(r);} if(!vis[l]){vis[l] = 1;dfs(l);} ans[++cnt] = t; } int main(){ Sca(N); if(N & 1){puts("-1"); return 0;} dfs(0); for(int i = cnt ; i >= 1; i --)cout << ans[i] << " " ; return 0; }
G.由于每天加的钱为实数而不要求为浮点数,一个显然的贪心是每两个取的物品i,j之间,每天加入的钱都是wj / (j - i)
dp是显然的,第一个难点在于每天钱数不增的限制,如果dp存储加入的钱数,2000 * 1e6很显然时间复杂度上过不去
一个比较巧妙地思想是dp[i][j]表示上一个操作是i - > j的物品的选择,2000 * 2000满足了时间复杂度还满足了钱数
得到状态转移方程dp[j][k] = max(dp[i][j] + V[k])
到了这一步就可以写出一个n3 的暴力(雾),将状态转移方程变形,得到i > j - W(j) / W(k) * (k - j)这样一个右边和i无关的方程.
所以考虑枚举j和k,然后就可以得到i的下界,i的上界显然为j - 1,就变成了一个后缀最大值的问题,这个甚至不需要树状数组,直接维护一个简单的一维数组即可.
时间复杂度n²
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x); #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x); #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double eps = 1e-6; const int maxn = 2010; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N; struct Good{ LL v; double w; }P[maxn]; LL dp[maxn][maxn]; LL Max[maxn]; int main(){ Sca(N); for(int i = 1; i <= N ; i ++) scanf("%lf",&P[i].w); for(int i = 1; i <= N ; i ++) Scl(P[i].v); for(int i = 0; i <= N ; i ++){ for(int j = 0; j <= N ; j ++){ dp[i][j] = -1e18; } } for(int i = 1; i <= N ; i ++) dp[0][i] = P[i].v; for(int j = 1; j <= N ; j ++){ Max[j] = -1e18; for(int k = j - 1; k >= 0; k --) Max[k] = max(Max[k + 1],dp[k][j]); for(int k = j + 1; k <= N ; k ++){ double l; if(!P[k].w) l = 0; else l = max(j - P[j].w / P[k].w * (k - j),(double)0); int L = (int)l; if(fabs(l - L) > eps) L++; if(L > j - 1) continue; dp[j][k] = Max[L] + P[k].v; // Max[k][j] } } LL ans = 0; for(int i = 0; i <= N ; i ++){ for(int j = i + 1; j <= N ; j ++){ ans = max(ans,dp[i][j]); } } Prl(ans); return 0; }
H.取个log就会发现变成AjlogAi > AilogAj,直接sort一波,特判一下前后相等的情况即可.
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x); #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x); #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 1e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; PIL a[maxn]; int ans[maxn]; bool cmp(PIL x,PIL y){ return 1.0 * x.se * log(y.se * 1.0) > 1.0 * y.se * log(x.se * 1.0); } bool equal(double x,double y){ return fabs(x - y) < eps; } int main(){ N = read(); for(int i = 1; i <= N ; i ++){ Scl(a[i].se); a[i].fi = i; } sort(a + 1,a + 1 + N,cmp); int cnt = 0; for(int i = 1; i <= N ; i ++){ if(i > 1 && equal(1.0 * a[i].se * log(a[i - 1].se * 1.0),1.0 * a[i - 1].se * log(a[i].se * 1.0))) cnt++; else cnt = 0; ans[a[i].fi] = i - 1 - cnt; } for(int i = 1; i <= N ; i ++){ printf("%d ",ans[i]); } return 0; }
I.要有多坑有多坑,正确题意为双射 + 最后25字母可推26字母,其他没有难度
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x); #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x); #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 1e6 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; int to[30],to2[30]; char str[maxn],str2[maxn]; int main(){ while(~scanf("%s%s",str,str2)){ int l = strlen(str); for(int i = 0 ; i < 26; i ++) to[i] = to2[i] = -1; for(int i = 0; str[i]; i ++){ int id1 = str[i] - 'a',id2 = str2[i] - 'a'; if(~to[id1] && to[id1] != id2){ puts("Impossible"); exit(0); } if(~to2[id2] && to2[id2] != id1){ puts("Impossible"); exit(0); } to[id1] = id2; to2[id2] = id1; } int cnt = 0; for(int i = 0 ; i < 26; i ++) if(~to[i]) cnt++; if(cnt == 25){ int t = 0; for(int i = 0 ; i < 26; i ++){ if(to[i] == -1) t = i; } for(int i = 0 ; i < 26; i ++) if(to2[i] == -1) to[t] = i; } for(int i = 0 ; i < 26; i ++){ if(~to[i]) printf("%c->%c\n",i + 'a',to[i] + 'a'); } } return 0; }
J.过于温暖,给出题人点赞
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x); #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x); #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 110; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; LL N,M,K,T; int main(){ cin >> N >> K >> T; cout << max(0LL,N - K * T) << endl; return 0; }
K.很显然是预处理出所有情况然后二分端点.
坑点1.不能用海伦公式,过不了double浮点数的误差,需要用向量叉积并且不除2,后面查询的时候将l和r乘2达到无浮点数的目的
2.r用upper_bound,l用lower_bound,最后r - l即可,天知道为什么我一开始手写了两个二分
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x); #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x); #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 310; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,Q; struct Point{ LL x,y; }point[maxn]; LL P[255 * 255 * 50]; LL cul(LL x1,LL y1,LL x2,LL y2){ //cout << x1 << " " << y1 << " " << x2 << " " << y2 << endl; return abs(x1 * y2 - x2 * y1); } int main(){ Sca2(N,Q); //x1:(ax - bx) y1:(ay - by) x2:ax - cx y2:ay - cy for(int i = 1; i <= N ; i ++) scanf("%lld%lld",&point[i].x,&point[i].y); int cnt = 0; for(int i = 1; i <= N ; i ++){ for(int j = i + 1; j <= N; j ++){ LL x1 = point[i].x - point[j].x,y1 = point[i].y - point[j].y; for(int k = j + 1; k <= N ; k ++){ LL x2 = point[i].x - point[k].x,y2 = point[i].y - point[k].y; P[++cnt] = cul(x1,y1,x2,y2); } } } sort(P + 1,P + 1 + cnt); //for(int i = 1; i <= cnt; i ++) cout << P[i] << " "; //cout << endl; for(int i = 1; i <= Q; i ++){ LL l,r; scanf("%lld%lld",&l,&r); l *= 2; r *= 2; l = lower_bound(P + 1,P + 1 + cnt,l) - P; r = upper_bound(P + 1,P + 1 + cnt,r) - P; //cout << l << ' ' << r << endl; Pri(r - l); } return 0; }
L.考虑从T开始跑一颗最短路树，那么如果当人在i点的时候去掉边，显然去掉非树边上的边的行为是无意义的，如果去掉了树边上的边，那么就需要走到他们的子树上某个结点j，通过一个非树边走到一个非子树的结点k，然后再到T点，距离就是dis[j] - dis[i] + edge(j,k) + dis[k]
显然（雾）我们可以对每一条非树边进行预处理，每一条连接(i,j)的非树边将会对i到lca(i,j),j到lca(i,j)的两条链产生贡献（目的是寻找一条i点树边被封之后最短的走到T的路线),这个过程可以用树链剖分实现
最后再从T开始往S跑最短路，由于我们已经求出了每个点i的树边被封之后到T的最短路，所以更新答案是dis[v] = max(dis[u.pos] + edge[i].dis,dis2[v]);
也就是说，如果在v点直接切断会比在之后切断造成需要走更长的路。
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x); #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x); #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();} while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;} const double eps = 1e-9; const int maxn = 2e5 + 10; const int maxm = 2e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,S,T; struct Edge{ int to,next; LL dis; }edge[maxm * 2]; struct E{ int u,v,flag; LL w; E(){} E(int u,int v,LL w):u(u),v(v),w(w){} }e[maxm * 2]; int head[maxn],tot; void init(){ for(int i = 0 ; i <= N ; i ++) head[i] = -1; tot = 0; } void add(int u,int v,LL w){ edge[tot].to = v; edge[tot].next = head[u]; edge[tot].dis = w; head[u] = tot++; } struct node{ int pos; LL cost; node(){} node(int pos,LL cost):pos(pos),cost(cost){} friend bool operator < (node a,node b){ return a.cost > b.cost; } }; LL dis[maxn],dis2[maxn]; int fa[maxn]; void Dijkstra(int s){ for(int i = 1; i <= N ; i ++) dis[i] = 1e18; priority_queue<node>Q; Q.push(node(s,0)); fa[s] = dis[s] = 0; while(!Q.empty()){ node u = Q.top(); Q.pop(); if(u.cost > dis[u.pos]) continue; for(int i = head[u.pos]; ~i; i = edge[i].next){ int v = edge[i].to; if(dis[v] > dis[u.pos] + edge[i].dis){ dis[v] = dis[u.pos] + edge[i].dis; fa[v] = u.pos; Q.push(node(v,dis[v])); } } } } int dep[maxn],top[maxn],pos[maxn],size[maxn],son[maxn]; void dfs1(int t,int la){ size[t] = 1; son[t] = t; int heavy = 0; for(int i = head[t]; ~i ; i = edge[i].next){ int v = edge[i].to; if(v == la) continue; dep[v] = dep[t] + 1; fa[v] = t; dfs1(v,t); if(size[v] > heavy){ heavy = size[v]; son[t] = v; } size[t] += size[v]; } } int cnt; void dfs2(int t,int la){ top[t] = la; pos[t] = ++cnt; if(son[t] == t) return; dfs2(son[t],la); for(int i = head[t]; ~i ; i = edge[i].next){ int v = edge[i].to; if((fa[t] == v) || v == son[t]) continue; dfs2(v,v); } } int lca(int u,int v){ while(top[u] != top[v]){ if(dep[top[u]] < dep[top[v]]) swap(u,v); u = fa[top[u]]; } if(dep[u] < dep[v]) return u; return v; } struct Tree{ int l,r; LL lazy; }tree[maxn << 2]; void Build(int t,int l,int r){ tree[t].l = l; tree[t].r = r; tree[t].lazy = 1e18; if(l == r) return; int m = l + r >> 1; Build(t << 1,l,m); Build(t << 1 | 1,m + 1,r); } void Pushdown(int t){ tree[t << 1].lazy = min(tree[t << 1].lazy,tree[t].lazy); tree[t << 1 | 1].lazy = min(tree[t << 1 | 1].lazy,tree[t].lazy); } void update(int t,int l,int r,LL sum){ if(l <= tree[t].l && tree[t].r <= r){ tree[t].lazy = min(tree[t].lazy,sum); return ; } Pushdown(t); int m = tree[t].l + tree[t].r >> 1; if(r <= m) update(t << 1,l,r,sum); else if(l > m) update(t << 1 | 1,l,r,sum); else{ update(t << 1,l,m,sum); update(t << 1 | 1,m + 1,r,sum); } } LL query(int t,int p){ if(tree[t].l == tree[t].r) return tree[t].lazy; Pushdown(t); int m = tree[t].l + tree[t].r >> 1; if(p <= m) return query(t << 1,p); else return query(t << 1 | 1,p); } void update(int u,int v,LL sum){ while(top[u] != top[v]){ update(1,pos[top[u]],pos[u],sum); u = fa[top[u]]; } if(u == v) return; update(1,pos[v] + 1,pos[u],sum); } int main(){ scanf("%d%d%d%d",&N,&M,&S,&T); init(); for(int i = 1; i <= M ; i ++){ int u,v; Sca2(u,v); LL w; Scl(w); add(u,v,w); add(v,u,w); e[i] = E(u,v,w); e[i].flag = 1; } Dijkstra(T); init(); for(int i = 1; i <= M ; i ++){ if(fa[e[i].u] == e[i].v || fa[e[i].v] == e[i].u){ add(e[i].v,e[i].u,e[i].w); add(e[i].u,e[i].v,e[i].w); }else e[i].flag = 0; } dfs1(T,T); dfs2(T,T); Build(1,1,N); for(int i = 1; i <= M ; i ++){ if(e[i].flag) continue; LL sum = dis[e[i].u] + dis[e[i].v] + e[i].w; int l = lca(e[i].u,e[i].v); update(e[i].u,l,sum); update(e[i].v,l,sum); } for(int i = 1; i <= N; i ++) dis2[i] = query(1,pos[i]) - dis[i]; for(int i = 1; i <= N ; i ++) dis[i] = 1e18; init(); for(int i = 1; i <= M ; i ++){ add(e[i].u,e[i].v,e[i].w); add(e[i].v,e[i].u,e[i].w); } priority_queue<node>Q; Q.push(node(T,0)); dis[T] = 0; while(!Q.empty()){ node u = Q.top(); Q.pop(); if(dis[u.pos] < u.cost) continue; for(int i = head[u.pos]; ~i ; i = edge[i].next){ int v = edge[i].to; if(dis[v] > max(dis[u.pos] + edge[i].dis,dis2[v])){ dis[v] = max(dis[u.pos] + edge[i].dis,dis2[v]); Q.push(node(v,dis[v])); } } } if(dis[S] >= 1e16) puts("-1"); else Prl(dis[S]); return 0; }
M.用SG函数打表博弈，反正我是不会做，贴上队友代码
#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> #include <map> #include <set> #include <iostream> #include <string> #include <algorithm> #include <vector> #include <queue> using namespace std; typedef long long LL; const int maxn = 1e6+500; int if_pre[maxn +5]; void pre_first() { memset(if_pre, 0x3f, sizeof(if_pre)); if_pre[0] = 0; if_pre[1] = 1; int pre_num =2; for (int i = 2; i < maxn; i++) { if (if_pre[i] == 0x3f3f3f3f) { if_pre[i] = pre_num++; for (int j = i << 1; j < maxn; j += i) { if_pre[j] = min(if_pre[j],if_pre[i]); } } } } int main() { int t; cin >>t; pre_first(); while (t --) { int n,a; cin >>n; int sum =0; while (n--) { cin >>a; sum^=if_pre[a]; //cout<<a<<'*'<<if_pre[a]<<endl; } if (sum) cout<<"Subconscious is our king!"<<endl; else cout<<"Long live with King Johann!"<<endl; } #ifdef VSCode system("pause"); #endif return 0; }
