hdu2507二分图 + 打印路径

来源：互联网发布：电视盒子软件应用破解编辑：程序博客网时间：2024/05/16 16:14

思路：对每个格子进行编号，然后在标记哪些格子是可以链接的，人后就是Hungary()的事情了；

这个应该是特判的；

/*****************************************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)#define pk push_backtemplate<class T> inline T Get_Max(const T&a,const T&b){return a < b?b:a;}template<class T> inline T Get_Min(const T&a,const T&b){return a < b?a:b;}typedef long long ll;typedef pair<int,int> ii;const int inf = 1 << 30;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int maxn = 10010;int link[maxn];int vis[maxn];int ok[maxn];int n,m;int total;bool dfs(int u){int k;k = u + 1;if (u % m + 1 < m && k < total && ok[k] != -1 && !vis[k]){vis[k] = 1;if (link[k] == -1 || dfs(link[k])){link[k] = u;return true;;}}k = u + m;if (k < total && ok[k] != -1 && !vis[k]){vis[k] = 1;if (link[k] == -1 || dfs(link[k])){link[k] = u;return true;}}k = u - 1;if (u % m > 0 && k >= 0 && ok[k] != -1 && !vis[k]){vis[k] = 1;if (link[k] == -1 || dfs(link[k])){link[k] = u;return true;}}k = u - m;if (k >= 0 && ok[k] != -1 && !vis[k]){vis[k] = 1;if (link[k] == -1 || dfs(link[k])){link[k] = u;return true;}}return false;}inline int Hungary(){int ret = 0;for (int i = 0;i < total;++i){if (ok[i] != -1){MEM(vis, 0);if (dfs(i)) ret++;}}return ret;}int main(){// ios::sync_with_stdio(false);// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);while(~scanf("%d%d",&n,&m)){if (n == 0 && m == 0) break;MEM(link, -1);total = n * m;for (int i = 0;i < total;i++)ok[i] = i;int x,y;int t;scanf("%d",&t);while(t--){scanf("%d%d",&x,&y);int tmp = (x - 1) * m + (y - 1);ok[tmp] = -1;// cout << "ok\n";}int ans = Hungary();// cout << ans << endl;printf("%d\n",ans / 2);MEM(vis, 0);for (int i = 0;i < total;i++){if (ok[i] != -1 && link[i] != -1  && !vis[link[i]]&& !vis[i]){vis[link[i]] = vis[i] = 1;printf("(%d,%d)--(%d,%d)\n",i/m + 1,i%m + 1,link[i]/m + 1,link[i]%m + 1);}}}return 0;}

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