UVA11419 SAM I AM

题目大意：给你一个n*m的矩阵，上面有一些格子上有目标，我们可以在格子的外面用枪打目标，一发子弹可以消灭一行或者一列目标，问你最少多少枪能把目标打光，并且输出开枪的位置（转自http://blog.csdn.net/u013761036/article/details/40154095）

显然的最小点覆盖。最开始用网络流做。。。发现整天TLE。。。发现输出方案有点瑕疵。。改了之后还是T。。看了看题。。

100,000条边？？？？？？？？

果断改匈牙利

一遍过。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
#include <string> 
#include <cmath> 
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
template<class T>
inline void swap(T &a, T &b)
{
    T tmp = a;a = b;b = tmp;
}
inline void read(int &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}
const int INF = 0x3f3f3f3f;
const int MAXN = 2000 + 10;
const int MAXM = 2000000 + 10;

int g[MAXN][MAXN], lk1[MAXN], lk2[MAXN], vis[MAXN], tmp1, tmp2, n, m, num;

int dfs(int u)
{
    for(int v = 1;v <= n;++ v)
    {
        if(!g[u][v] || vis[v]) continue;
        vis[v] = 1;
        if(lk2[v] == -1 || dfs(lk2[v]))
        {
            lk2[v] = u;
            lk1[u] = v;
            return 1;
        }
    }
    return 0;
}

int xiongyali()
{
    int ans = 0;
    memset(lk1, -1, sizeof(lk1)), memset(lk2, -1, sizeof(lk2));
    for(int i = 1;i <= n;++ i)
    {
        memset(vis, 0, sizeof(vis));
        ans += dfs(i);
    }
    return ans;
}

int vis1[MAXN], vis2[MAXN];
void ddfs(int u)
{
    vis1[u] = 1;
    for(int v = 1;v <= n;++ v)
    {
        if(lk2[v] == -1 || !g[u][v] || vis2[v] || vis1[lk2[v]]) continue;
        vis2[v] = 1;
        if(lk2[v] == -1) continue;
        ddfs(lk2[v]);
    }
}

int main()
{
    while(scanf("%d %d %d", &n, &m, &num) != EOF && n && m && num)
    {
        memset(g, 0, sizeof(g)), memset(vis1, 0, sizeof(vis1)), memset(vis2, 0, sizeof(vis2));
        for(int i = 1;i <= num;++ i) 
            read(tmp1), read(tmp2), g[tmp1][tmp2] = 1;
        printf("%d ", xiongyali());
        for(int i = 1;i <= n;++ i)
            if(lk1[i] == -1) ddfs(i);
        for(int i = 1;i <= n;++ i) if(!vis1[i]) printf("r%d ", i);
        for(int i = 1;i <= m;++ i) if(vis2[i]) printf("c%d ", i);
        putchar('
');
    }
    return 0;
}