标签:不同的 href template pac its get lin put div
最直观的想法是贪心取, 然后网络流取check可不可行, 然后T了。
想到最大流可以等于最小割, 那么我们状压枚举字符代表的6个点连向汇点是否断掉,
然后再枚举64个本质不同的位置, 是否需要切段原点联想它的边, 复杂度电磁check复杂度64 * 64
用sosdp能优化到64 * 6
#include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ALL(x) (x).begin(), (x).end() using namespace std; const int N = 2e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;} int n, m, mask[N], c[N], cnt[N], sos[N]; char s[N], t[N], ans[N]; bool check() { int sum = 0, ret = inf; for(int i = 0; i < 6; i++) sum += c[i]; for(int i = 0; i < 64; i++) sos[i] = cnt[i]; for(int i = 0; i < 6; i++) for(int j = 0; j < 64; j++) if(j >> i & 1) sos[j] += sos[j ^ (1 << i)]; for(int s1 = 0; s1 < 64; s1++) { int tmp = 0; for(int i = 0; i < 6; i++) if(s1 >> i & 1) tmp += c[i]; tmp += sum - sos[s1]; chkmin(ret, tmp); } return sum == ret; } inline int getId(char c) { return c - ‘a‘; } int main() { scanf("%s", s + 1); n = strlen(s + 1); for(int i = 1; i <= n; i++) c[getId(s[i])]++; scanf("%d", &m); while(m--) { int p; scanf("%d%s", &p, t + 1); for(int i = 1; t[i]; i++) mask[p] |= 1 << getId(t[i]); } for(int i = 1; i <= n; i++) { if(!mask[i]) mask[i] = 63; cnt[mask[i]]++; } for(int i = 1; i <= n; i++) { bool flag = false; for(int j = 0; j < 6; j++) { if(c[j] && mask[i] >> j & 1) { c[j]--; cnt[mask[i]]--; flag = check(); if(flag) { ans[i] = ‘a‘ + j; break; } c[j]++; cnt[mask[i]]++; } } if(!flag) return puts("Impossible"), 0; } ans[n + 1] = ‘\0‘; puts(ans + 1); return 0; } /* */
Codeforces 1009G Allowed Letters 最大流转最小割 sosdp
标签:不同的 href template pac its get lin put div
原文地址:https://www.cnblogs.com/CJLHY/p/10727182.html