[BC冠军赛(online)]小结
A Movie
题意：给你n个区间，判断能否选出3个不相交的区间。
思路：令f(i)表示能否选出两个不相交区间并且以区间i为右区间的值，g(i)表示能否选出两个不相交区间并且以区间i为左区间的值，如果存在i，f(i) && g(i）== true，则存在这样的3个不相交区间。计算f数组的时候，只要从前往后和从后往前各扫一遍，表示对于i而言，另一个区间来自于它的前面（后面），同时维护右边界的最小值（因为只需关心前面（后面）的所有区间的右边界的最小值）。对于g数组类似处理。
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 #include <bitset> 17 #include <functional> 18 #include <numeric> 19 #include <stdexcept> 20 #include <utility> 21 22 using namespace std; 23 24 #define mem0(a) memset(a, 0, sizeof(a)) 25 #define mem_1(a) memset(a, -1, sizeof(a)) 26 #define lson l, m, rt << 1 27 #define rson m + 1, r, rt << 1 | 1 28 #define define_m int m = (l + r) >> 1 29 #define rep_up0(a, b) for (int a = 0; a < (b); a++) 30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++) 31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) 32 #define rep_down1(a, b) for (int a = b; a > 0; a--) 33 #define all(a) (a).begin(), (a).end() 34 #define lowbit(x) ((x) & (-(x))) 35 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 36 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 37 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 38 #define pchr(a) putchar(a) 39 #define pstr(a) printf("%s", a) 40 #define sstr(a) scanf("%s", a) 41 #define sint(a) scanf("%d", &a) 42 #define sint2(a, b) scanf("%d%d", &a, &b) 43 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c) 44 #define pint(a) printf("%d\n", a) 45 #define test_print1(a) cout << "var1 = " << a << endl 46 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl 47 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl 48 49 typedef long long LL; 50 typedef pair<int, int> pii; 51 typedef vector<int> vi; 52 53 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1}; 54 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 }; 55 const int maxn = 1e7 + 7; 56 const int md = 10007; 57 const int inf = 1e9 + 7; 58 const LL inf_L = 1e18 + 7; 59 const double pi = acos(-1.0); 60 const double eps = 1e-6; 61 62 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 63 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;} 64 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;} 65 template<class T>T condition(bool f, T a, T b){return f?a:b;} 66 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} 67 int make_id(int x, int y, int n) { return x * n + y; } 68 69 struct Node { 70 unsigned int l, r; 71 bool operator < (const Node &that) const { 72 return l < that.l || l == that.l && r < that.r; 73 } 74 }; 75 Node node[maxn]; 76 unsigned int n, l, r, a, b, c, d; 77 bool f1[maxn], f2[maxn], g1[maxn], g2[maxn]; 78 int main() { 79 //freopen("in.txt", "r", stdin); 80 int T; 81 cin >> T; 82 while (T--) { 83 cin >> n >> l >> r >> a >> b >> c >> d; 84 node[1].l = l; 85 node[1].r = r; 86 for (int i = 2; i <= n; i++) { 87 node[i].l = node[i - 1].l * a + b; 88 node[i].r = node[i - 1].r * c + d; 89 } 90 rep_up1(i, n) { 91 if (node[i].l > node[i].r) swap(node[i].l, node[i].r); 92 } 93 mem0(f1); 94 mem0(f2); 95 mem0(g1); 96 mem0(g2); 97 unsigned int min_r = 0xffffffff; 98 rep_up1(i, n) { 99 f1[i] = node[i].l > min_r; 100 min_update(min_r, node[i].r); 101 } 102 min_r = 0xffffffff; 103 rep_down1(i, n) { 104 f2[i] = node[i].l > min_r; 105 min_update(min_r, node[i].r); 106 } 107 unsigned int max_l = 0; 108 rep_up1(i, n) { 109 g1[i] = node[i].r < max_l; 110 max_update(max_l, node[i].l); 111 } 112 max_l = 0; 113 rep_down1(i, n) { 114 g2[i] = node[i].r < max_l; 115 max_update(max_l, node[i].l); 116 } 117 bool ok = false; 118 rep_up1(i, n) { 119 ok = ok || (f1[i] || f2[i]) && (g1[i] || g2[i]); 120 if (ok) break; 121 } 122 puts(ok? "YES" : "NO"); 123 } 124 return 0; 125 }
题意：判断一个有有向边和无向边的图是否存在环。每条边只能走一次，可能有重边，但没有自环。
思路：对于无向边连接的点可以缩为一个点，原来的点上的有向边连到缩成的点上。实际上就是判断用缩成的点建成的有向图是否存在环。并查集+dfs即可。
1 #pragma comment(linker, "/STACK:1024000000,1024000000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 #include <bitset> 17 #include <functional> 18 #include <numeric> 19 #include <stdexcept> 20 #include <utility> 21 22 using namespace std; 23 24 #define mem0(a) memset(a, 0, sizeof(a)) 25 #define mem_1(a) memset(a, -1, sizeof(a)) 26 #define lson l, m, rt << 1 27 #define rson m + 1, r, rt << 1 | 1 28 #define define_m int m = (l + r) >> 1 29 #define rep_up0(a, b) for (int a = 0; a < (b); a++) 30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++) 31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) 32 #define rep_down1(a, b) for (int a = b; a > 0; a--) 33 #define all(a) (a).begin(), (a).end() 34 #define lowbit(x) ((x) & (-(x))) 35 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 36 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 37 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 38 #define pchr(a) putchar(a) 39 #define pstr(a) printf("%s", a) 40 #define sstr(a) scanf("%s", a) 41 #define sint(a) scanf("%d", &a) 42 #define sint2(a, b) scanf("%d%d", &a, &b) 43 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c) 44 #define pint(a) printf("%d\n", a) 45 #define test_print1(a) cout << "var1 = " << a << endl 46 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl 47 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl 48 49 typedef long long LL; 50 typedef pair<int, int> pii; 51 typedef vector<int> vi; 52 53 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1}; 54 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 }; 55 const int maxn = 1e6 + 7; 56 const int md = 10007; 57 const int inf = 1e9 + 7; 58 const LL inf_L = 1e18 + 7; 59 const double pi = acos(-1.0); 60 const double eps = 1e-6; 61 62 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 63 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;} 64 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;} 65 template<class T>T condition(bool f, T a, T b){return f?a:b;} 66 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} 67 int make_id(int x, int y, int n) { return x * n + y; } 68 69 struct UnionFindSets { 70 vi f; 71 int N; 72 UnionFindSets() { f.clear(); } 73 void clear() { f.clear(); } 74 void resize(int n) { N = n + 2; f.resize(n + 5); } 75 void Init() { for (int i = 1; i <= N; i++) f[i] = i; } 76 int get(int u) { if (u == f[u]) return u; return f[u] = get(f[u]); } 77 void add(int u, int v) { f[get(u)] = get(v); } 78 bool find(int u, int v) { return get(u) == get(v); } 79 }; 80 81 82 template<class edge> struct Graph { 83 vector<vector<edge> > adj; 84 Graph(int n) { adj.clear(); adj.resize(n + 5); } 85 Graph() { adj.clear(); } 86 void resize(int n) { adj.resize(n + 5); } 87 void add(int s, edge e){ adj[s].push_back(e); } 88 void del(int s, edge e) { adj[s].erase(find(iter(adj[s]), e)); } 89 void clear() { adj.clear(); } 90 vector<edge>& operator [](int t) { return adj[t]; } 91 }; 92 Graph<int> G; 93 UnionFindSets ufs; 94 int vis[maxn], mark[maxn]; 95 bool ok; 96 97 void dfs(int pos) { 98 vis[pos] = -1; 99 int sz = G[pos].size(); 100 rep_up0(i, sz) { 101 int u = G[pos][i]; 102 if (vis[u] == -1) { 103 ok = true; 104 return ; 105 } 106 if (!ok && !vis[u]) dfs(u); 107 } 108 vis[pos] = 1; 109 } 110 int main() { 111 //freopen("in.txt", "r", stdin); 112 int T, n, m1, m2; 113 cin >> T; 114 while (T--) { 115 cin >> n >> m1 >> m2; 116 G.clear(); 117 G.resize(n); 118 ufs.clear(); 119 ufs.resize(n); 120 ufs.Init(); 121 ok = false; 122 rep_up0(i, m1) { 123 int u, v; 124 sint2(u, v); 125 if (ufs.find(u, v)) { 126 ok = true; 127 } 128 ufs.add(u, v); 129 } 130 rep_up0(i, m2) { 131 int u, v; 132 sint2(u, v); 133 G.add(ufs.get(u), ufs.get(v)); 134 } 135 mem0(mark); 136 rep_up1(i, n) { 137 if (mark[ufs.get(i)]) continue; 138 mark[ufs.get(i)] = 1; 139 G.add(0, ufs.get(i)); 140 } 141 mem0(vis); 142 if (!ok) dfs(0); 143 puts(ok? "YES" : "NO"); 144 } 145 return 0; 146 }
J GCD
题意：给m个询问Li, Ri, Pi，表示[Li, Ri]的所有数的gcd等于Pi，求原来的数组（所有数的和小的优先）。
思路：对于数组中某个位置上的数a[x]，考虑所有覆盖位置x的区间，那么a[x]是所有这些区间的Pi的最小公倍数的倍数，这是显然的，因为a[x]必须是每个Pi的倍数，不妨先把a数组取个最小值，令a[x] = lcm(Pi)(Li<=x<=Ri)。另一方面，对于某一个区间[Li, Ri]而言，因为这里的a[Li]~a[Ri]都是Pi的倍数了，所以gcd(a[Li], a[Li + 1] ,..., a[Ri])>=Pi，而如果gcd(a[Li]~a[ri])>Pi，则无论怎样调整a数组的值（注意：调整只能整倍的放大）,gcd(a[Li]~R[i])始终大于Pi，故无解。如果gcd(a[Li]~a[Ri])=Pi了，a数组里的值就是答案，因为不能再小了。
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 #include <bitset> 17 #include <functional> 18 #include <numeric> 19 #include <stdexcept> 20 #include <utility> 21 22 using namespace std; 23 24 #define mem0(a) memset(a, 0, sizeof(a)) 25 #define mem_1(a) memset(a, -1, sizeof(a)) 26 #define lson l, m, rt << 1 27 #define rson m + 1, r, rt << 1 | 1 28 #define define_m int m = (l + r) >> 1 29 #define rep_up0(a, b) for (int a = 0; a < (b); a++) 30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++) 31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) 32 #define rep_down1(a, b) for (int a = b; a > 0; a--) 33 #define all(a) (a).begin(), (a).end() 34 #define lowbit(x) ((x) & (-(x))) 35 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 36 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 37 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 38 #define pchr(a) putchar(a) 39 #define pstr(a) printf("%s", a) 40 #define sstr(a) scanf("%s", a) 41 #define sint(a) scanf("%d", &a) 42 #define sint2(a, b) scanf("%d%d", &a, &b) 43 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c) 44 #define pint(a) printf("%d\n", a) 45 #define test_print1(a) cout << "var1 = " << a << endl 46 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl 47 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl 48 49 typedef long long LL; 50 typedef pair<int, int> pii; 51 typedef vector<int> vi; 52 53 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1}; 54 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 }; 55 const int maxn = 3e4 + 7; 56 const int md = 10007; 57 const int inf = 1e9 + 7; 58 const LL inf_L = 1e18 + 7; 59 const double pi = acos(-1.0); 60 const double eps = 1e-6; 61 62 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 63 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;} 64 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;} 65 template<class T>T condition(bool f, T a, T b){return f?a:b;} 66 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} 67 int make_id(int x, int y, int n) { return x * n + y; } 68 69 LL lcm(LL a, LL b) { 70 return a / gcd(a, b) * b; 71 } 72 int L[1007], R[1007], P[1007], a[1007]; 73 74 int main() { 75 //freopen("in.txt", "r", stdin); 76 int T, n, q; 77 cin >> T; 78 while (T--) { 79 cin >> n >> q; 80 rep_up0(i, q) { 81 sint3(L[i], R[i], P[i]); 82 } 83 bool ok = true; 84 rep_up1(i, n) { 85 LL x = 1; 86 rep_up0(j, q) { 87 if (L[j] <= i && R[j] >= i) { 88 x = lcm(x, P[j]); 89 if (x > 1e9) { 90 ok = false; 91 break; 92 } 93 } 94 } 95 a[i] = x; 96 if (!ok) break; 97 } 98 if (!ok) { 99 puts("Stupid BrotherK!"); 100 continue; 101 } 102 rep_up0(i, q) { 103 int x = a[L[i]]; 104 for (int j = L[i] + 1; j <= R[i]; j ++) { 105 x = gcd(x, a[j]); 106 } 107 if (x != P[i]) { 108 ok = false; 109 break; 110 } 111 } 112 if (!ok) { 113 puts("Stupid BrotherK!"); 114 continue; 115 } 116 rep_up1(i, n) { 117 printf("%d%c", a[i], i == n? '\n' : ' '); 118 } 119 } 120 return 0; 121 }
PS：
因为太弱，只看了这三题，其它题等题目挂出来尽量补上。
后来发现别人A题是暴力水过的，由于区间是随机的，所以当n很大的时候，基本可以断定答案为yes了。
----------------------------- UPDATE：5月8日 ----------------------------------
由于最近比较忙，自己又水平不够，补题进度太慢。。。。。
做了一下B,C,E，详见最近的随笔记录。
---------------------------- update: 5.9-----------------------------
补完短时间内可以出的D题，其它三个题暂时不想看了，以后有时间再慢慢搞吧=.=
