poj3698(二分图最大匹配 + 拆点)

来源：互联网发布：网络本科需要统考么编辑：程序博客网时间：2024/05/14 02:59

思路：把电影拆开，，，对于每部电影的去的天数呢分开，，(1...d1)(d1 + 1.....d1 + d2)....入上面这种拆点。

然后就是每周的周几可以去，就把周扩展成天：比如说1表示第一周的周一，8表示第二周的周一，如此类推，

用上下这种扩展的方式简图，，，只要最后的匹配数目等于总的天数，那么就是可以的。

点击题目链接

/*****************************************Author      :Crazy_AC(JamesQi)Time        :2015File Name   :*****************************************/// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <algorithm>#include <iomanip>#include <sstream>#include <string>#include <stack>#include <queue>#include <deque>#include <vector>#include <map>#include <set>#include <stdio.h>#include <string.h>#include <math.h>#include <stdlib.h>#include <limits.h>using namespace std;#define MEM(a,b) memset(a,b,sizeof a)typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> ii;const int inf = 1 << 30;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;int T,n;int d,w;int day[8];int g[1010][400];int vN,uN;int link[410];bool vis[410];bool dfs(int u){for (int i = 1;i <= vN;i++){if (g[u][i] && !vis[i]){vis[i] = true;if (link[i] == -1 || dfs(link[i])){link[i] = u;return true;}}}return false;}int Hungary(){int ret = 0;MEM(link, -1);for (int i = 1;i <= vN;i++){MEM(vis, false);if (dfs(i)) ret++;}return ret;}int main(){// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);cin >> T;while(T--){cin >> n;int st = 1,ed = 0;int sum = 0;uN = vN = 0;MEM(g, 0);MEM(day, 0);for (int i = 1;i <= n;i++){for (int j = 1;j <= 7;j++){scanf("%d",&day[j]);}scanf("%d%d",&d,&w);for (int j = st;j < st + d;j++){for (int k = 1;k <= 7;k++){for (int l = 0;l < w;l++){g[j][k + l * 7] = day[k];}}}st += d;sum += d;vN = max(vN,7 * w);}uN = st - 1;int ans = Hungary();if (ans == sum) puts("Yes");else puts("No");}return 0;}

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