codeforces-1140 (div2)
A.维护一个前缀最大值，不断跳即可
#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 = 1e4 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; int a[maxn]; int Max[maxn]; int main(){ Sca(N); for(int i = 1; i <= N ; i ++){ Sca(a[i]); a[i] = max(a[i],a[i - 1]); } int cnt = 0; int now = 0; while(now < N){ cnt++; now++; while(now != a[now]) now = a[now]; } Pri(cnt); return 0; }
B.从左边开始删除一直删到>或者从右边开始删除一直删到<，两种方案取最小值。
#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; int N,M,K; char str[maxn]; int main(){ int T = read(); while(T--){ Sca(N); scanf("%s",str + 1); int ans = N - 1; for(int i = 1; i <= N ; i ++){ if(str[i] == '>'){ ans = min(ans,i - 1); break; } } for(int i = N ; i >= 1; i --){ if(str[i] == '<'){ ans = min(ans,N - i); break; } } Pri(ans); } return 0; }
C.按照beauty值从大到小排序，维护一个前缀和（最多K个length）,小根堆维护一下每次要踢出哪个最小lenght的弟弟
#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 = 3e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; PLL P[maxn]; bool cmp(PLL a,PLL b){ return a.se > b.se; } int main(){ Sca2(N,K); for(int i = 1; i <= N; i ++){ scanf("%lld%lld",&P[i].fi,&P[i].se); } sort(P + 1,P + 1 + N,cmp); priority_queue<LL,vector<LL>,greater<LL>>Q; LL sum = 0; LL ans = 0; for(int i = 1; i <= N ; i ++){ ans = max(ans,(sum + P[i].fi) * P[i].se); Q.push(P[i].fi); sum += P[i].fi; if(Q.size() >= K){ sum -= Q.top(); Q.pop(); } } Prl(ans); return 0; }
D.发现1,2,3| 1,3,4| 1,4,5 .. |1,n - 1,n为最优解。
#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 = 3e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; int main(){ Sca(N); LL sum = 0; for(int i = 2; i < N ; i ++){ sum += i * (i + 1); } Prl(sum); return 0; }
E*
1.发现不存在奇长度的回文串只需要满足不存在长度3的回文串即可，长度为5的回文串包含了长度3的回文串。
2.也就是说，要满足所有的str[x] != str[x + 2],发现奇偶位置上的数字其实互不干扰，那可以把序列分成两个序列来做。
3.得到dp方程,dp[i][j]表示到了i这个位置，且这个位置上的数字为j的时候，满足条件序列的总数，状态转移方程为dp[i][j] = ∑dp[i - 1][p] (1 <= p <= K && p != j)
4.当然这个时间复杂度和空间复杂度双双nk的算法是行不通的，我们发现对于i相同的dp[i][j],只会有两种不同的值，并且是其中k - 1个值相同以及另一个值鹤立鸡群的情况（当然也有可能都是鸡完全相同）
5.考虑标注特殊的点位置和特殊的点值以及k - 1个其他的点值，即可压缩复杂度为O(n)
#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 = 998244353; int N,M,K; LL a[maxn],b[maxn]; LL solve(LL *x,int n){ LL big,small,pos; if(x[1] == -1){ big = 1; small = 1; pos = 1; }else{ big = 0; small = 1; pos = x[1]; } for(int i = 2; i <= n ; i ++){ if(x[i] == -1){ LL s = big * (K - 1),b = big * (K - 2) + small; small = s % mod; big = b % mod; }else{ if(pos == x[i]){ small = big * (K - 1) % mod; big = 0; }else{ small = big * (K - 2) + small; small %= mod; pos = x[i]; big = 0; } } } return (small + big * (K - 1)) % mod; } int main(){ Sca2(N,K); int cnt1 = 0,cnt2 = 0; for(int i = 1; i <= N ; i ++){ if(i & 1) a[++cnt1] = read(); else b[++cnt2] = read(); } Prl(solve(a,cnt1) * solve(b,cnt2) % mod); return 0; }
F*
1.考虑将两个集合用一个二维的表来表示，添加就是将i行j列合并，最终答案是每个联通块的行数 * 列数。
2.考虑到用并查集,size1[maxn],size2[maxn],分别表示这个集合里面行列的数量。
3.问题在于删除，并查集是不存在在线直接删除这种黑科技的，但是我们可以用离线的方式，将并查集的存在生命周期放到线段树上，然后dfs整颗线段树进行优化。
这个操作在BZOJ2049里遇到过了，当时还写了blog https://www.cnblogs.com/Hugh-Locke/p/10367480.html
结果这次还是翻皮水了（雾）
#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 = 6e5 + 10; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; map<PII,int>P; //并查集 int fa[maxn]; LL size1[maxn],size2[maxn]; int dep[maxn]; void init(){ for(int i = 0; i < maxn; i ++){ fa[i] = i; if(i > 3e5) size2[i] = 1; else size1[i] = 1; dep[i] = 0; } } //线段树 struct Edge{ PII to; int next; }edge[maxn * 20]; int tot; void add(int& t,PII w){ edge[tot].to = w; edge[tot].next = t; t = tot++; } struct Tree{ int l,r; int head; }tree[maxn << 2]; void Build(int t,int l,int r){ tree[t].l = l; tree[t].r = r; tree[t].head = -1; if(l == r) return; int m = l + r >> 1; Build(t << 1,l,m); Build(t << 1 | 1,m + 1,r); } void add(int t,int l,int r,PII w){ if(l <= tree[t].l && tree[t].r <= r){ add(tree[t].head,w); return; } int m = tree[t].l + tree[t].r >> 1; if(r <= m) add(t << 1,l,r,w); else if(l > m) add(t << 1 | 1,l,r,w); else{ add(t << 1,l,m,w); add(t << 1 | 1,m + 1,r,w); } } int Stack[maxn],top; int find(int x){ while(fa[x] != x) x = fa[x]; return x; } LL ans; void rewind(int t){ while(top > t){ int x = Stack[--top]; dep[fa[x]] -= dep[x] + 1; ans -= 1LL * size1[fa[x]] * size2[fa[x]]; size1[fa[x]] -= size1[x]; size2[fa[x]] -= size2[x]; ans += 1LL * size1[fa[x]] * size2[fa[x]]; ans += 1LL * size1[x] * size2[x]; fa[x] = x; } } void dfs(int t){ int now = top; for(int i = tree[t].head; ~i ; i = edge[i].next){ PII t = edge[i].to; t.fi = find(t.fi); t.se = find(t.se); if(t.fi == t.se) continue; if(dep[t.fi] > dep[t.se]) swap(t.fi,t.se); Stack[top++] = t.fi; fa[t.fi] = t.se; dep[t.se] += dep[t.fi] + 1; ans -= 1LL * size1[t.fi] * size2[t.fi]; ans -= size1[t.se] * size2[t.se]; size1[t.se] += size1[t.fi]; size2[t.se] += size2[t.fi]; ans += 1LL * size1[t.se] * size2[t.se]; } if(tree[t].l == tree[t].r){ printf("%lld ",ans); }else{ dfs(t << 1); dfs(t << 1 | 1); } rewind(now); } int main(){ Sca(N); init(); Build(1,1,N); for(int i = 1; i <= N; i ++){ PII p; p.fi = read(),p.se = read() + 3e5; if(P[p]){ add(1,P[p],i - 1,p); P[p] = 0; }else P[p] = i; } for(map<PII,int>::iterator it = P.begin(); it != P.end(); it++) if((*it).se) add(1,(*it).se,N,(*it).fi); dfs(1); return 0; }
