uva 11419（最大匹配，最小点覆盖）

Problem C

SAM I AM

Input: Standard Input
Output: Standard Output
The world is in great danger!! Mental's forces have returned to Earth to eradicate humankind. Our last hope to stop this great evil is
Sam “Serious” Stone. Equipped with various powerful weapons, Serious Sam starts his mission to destroy the forces of evil.

After fighting two days and three nights, Sam is now in front of the temple KOPTOS where Mental's general Ugh Zan III is waiting for him. But this time, he has a serious problem. He is in shortage of ammo and a lot of enemies crawling inside the temple waiting
for him. After rounding the temple Sam finds that the temple is in rectangle shape and he has the locations of all enemies in the temple.

All of a sudden he realizes that he can kill the enemies without entering the temple using the great cannon ball which spits out a gigantic ball bigger than him killing anything it runs into and keeps on rolling until it finally explodes. But the cannonball
can only shoot horizontally or vertically and all the enemies along the path of that cannon ball will be killed.

Now he wants to save as many cannon balls as possible for fighting with Mental. So, he wants to know the minimum number of cannon balls and the positions from which he can shoot the cannonballs to eliminate all enemies from outside that temple.

Input

Here, the temple is defined as a RXC grid. The first line of each test case contains 3 integers: R(0<R<1001), C(0<C<1001) representing the grid of temple (R means number of row and C means number of column of the grid) and the number of enemies N(0<N<1000001)
inside the temple. After that there are N lines each of which contains 2 integers representing the position of the enemies in that temple. Each test case is followed by a new line (except the last one). Input is terminated when R=C=N=0. The size of the input
file is around 1.3 MB.

Output

For each test case there will be one line output. First print the minimum number (m) of cannonballs needed to wipe out the enemies followed by a single space and then
m positions from which he can shoot those cannonballs. For shooting horizontally print “r” followed by the row number and for vertical shooting print “c” followed by the column number. If there is
more than one solution any one will do.

Sample Input Output for Sample Input

4 4 3 1 1 1 4 3 2 4 4 2 1 1 2 2 0 0 0

2 r1 r3 2 r1 r2

Problemsetter: Syed Monowar Hossain

Special Thanks: Derek Kisman

本题是道经典的最小点覆盖模型的题目，在棋盘格中这种分行列的建图方法还是很常见的。

然后就是要用到定理：最小覆盖数等于最大匹配数。（最小点覆盖，即选择最少的点，使得每条边至少有一个端点被选中）

简单证明如下：1.最大匹配数>=最小覆盖数，假设已经求得了最大匹配，而在最大匹配选择了的边之外，还存在一条边的两个端点都没被选中，则这条边一定会在求最大匹配时被选中，与假设矛盾

2.最大匹配数<=最小覆盖数，因为最小点覆盖就是要求每条边至少有一个端点被选中，所以最大匹配时被选的那些边，都至少要有一个点被覆盖。

由1，2可得证定理。

构造最小覆盖的解书上有讲述，也没花时间去想，直接套了模版。

#include <cstdio>
#include <string>
#include <cstring>
#include <queue>
#include <algorithm>
#include <iostream>
using namespace std;

const int maxn = 1000 + 5;

struct BPM{
    int n,m;
    vector<int > G[maxn];
    int left[maxn];
    bool T[maxn];

    int right[maxn];
    bool S[maxn];

    void init(int n,int m){
        this -> n = n;
        this -> m = m;
        for(int i = 0;i < maxn;i++) G[i].clear();
    }

    void AddEdge(int u,int v){
        G[u].push_back(v);
    }

    bool match(int u){
        S[u] = true;
        for(int i = 0;i < G[u].size();i++){
            int v = G[u][i];
            if(!T[v]){
                T[v] = true;
                if(left[v] == -1 || match(left[v])){
                    left[v] = u;
                    right[u] = v;
                    return true;
                }
            }
        }
        return false;
    }

    int solve(){
        memset(left,-1,sizeof(left));
        memset(right,-1,sizeof(right));
        int ans = 0;
        for(int u = 0;u < n;u++){
            memset(S,0,sizeof(S));
            memset(T,0,sizeof(T));
            if(match(u))    ans++;
        }
        return ans;
    }

    int mincover(vector<int>& X,vector<int>& Y){
        int ans = solve();
        memset(S,0,sizeof(S));
        memset(T,0,sizeof(T));
        for(int u = 0;u < n;u++)
            if(right[u] == -1)  match(u);
        for(int u = 0;u < n;u++)
            if(!S[u])   X.push_back(u);
        for(int v = 0;v < n;v++)
            if(T[v])   Y.push_back(v);
        return  ans;
    }
};

BPM solver;

int main(){
    int n,r,c;

    while(scanf("%d%d%d",&r,&c,&n)){
        if(r == 0 && c == 0 && n == 0)  break;
        solver.init(r,c);
        while(n--){
            int x,y;
            scanf("%d%d",&x,&y);x--;y--;//模版中编号是从0开始的
            solver.AddEdge(x,y);
        }
        vector<int> X,Y;
        printf("%d",solver.mincover(X,Y));
        for(int i = 0;i < X.size();i++)
            printf(" r%d",X[i]+1);//最后别忘了加回来
        for(int j = 0;j < Y.size();j++)
            printf(" c%d",Y[j]+1);
        printf("\n");
    }
    return 0;
}