HDU-3360 National Treasures

给定各个宝物的多个守卫点，求最少移走宝物个数。

假如某个宝物的守卫点位于另一个宝物的位置上，则两者只能选其一，于是连边，求二分图的最小顶点覆盖。（可证明图内无奇数环）

//#include <cmath>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <fstream>
#include <iostream>

#define rep(i, l, r) for(int i=l; i<=r; i++)
#define clr(x, c) memset(x, c, sizeof(x))
#define N 53
#define MAX 1<<30
#define ll long long

using namespace std;
int read()
{
	int x=0, f=1; char ch=getchar();
	while (ch<'0' || ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	while (ch>='0' && ch<='9') { x=x*10+ch-'0'; ch=getchar(); }
	return x*f;
}

const int dir[12][2] = {{-1,-2},{-2,-1},{-2,+1},{-1,+2},{+1,+2},{+2,+1},{+2,-1},{+1,-2},{-1,0},{0,+1},{+1,0},{0,-1}};

struct edge{int y1, y2, n;} e[N*N*N*N]; int fir[N][N], en;
int n, m, k1[N][N], k2[N][N], t, ans, map[N][N];
bool b[N][N];

void Add(int x1, int x2, int y1, int y2) 
{ 
	en++, e[en].y1=y1, e[en].y2=y2, e[en].n=fir[x1][x2], fir[x1][x2]=en; 
	en++, e[en].y1=x1, e[en].y2=x2, e[en].n=fir[y1][y2], fir[y1][y2]=en; 
}

bool Can(int x, int y) { return x>0 && x<=n && y>0 && y<=m; }

bool Find(int x1, int x2)
{
	int o=fir[x1][x2], y1=e[o].y1, y2=e[o].y2;
	while (o) 
	{
		if (!b[y1][y2])
		{
			b[y1][y2]=1; 
			if (!k1[y1][y2] || Find(k1[y1][y2], k2[y1][y2])) { k1[y1][y2]=x1; k2[y1][y2]=x2; return true; }
		}
		o=e[o].n, y1=e[o].y1, y2=e[o].y2;
	}
	return false;
}

int main()
{
	while(scanf("%d %d", &n, &m) && (n || m))
	{
		t++; clr(fir, 0); en=ans=0;
		rep(i, 1, n) rep(j, 1, m) map[i][j]=read();
		rep(i, 1, n) rep(j, 1, m)
		{
			int a=map[i][j]; if (a<0) continue;
			rep(o, 0, 11)
			{
				if (a%2 && Can(i+dir[o][0], j+dir[o][1]) && map[i+dir[o][0]][j+dir[o][1]]>=0) 
					Add(i, j, i+dir[o][0], j+dir[o][1]); 
				a/=2;
			}
		}
		clr(k1, 0); clr(k2, 0);
		rep(i, 1, n) rep(j, 1, m) if ((i+j)%2)
		{
			clr(b, 0); if (Find(i, j)) ans++;
		}
		printf("%d. %d\n", t, ans);
	}
	return 0;
}