UVA 12549 （二分圖匹配）

對於這種x,y座標的，算是老套路了，每個點就是把x,y連在一起
選取最少點覆蓋所有的x,y，那麼就是個二分圖匹配，只是這題有障礙物，所以需要拆點，重新賦值行和列
思路代碼參考這篇文章

---

#include <iostream>
#include <algorithm>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <set>
#include <cmath>
#include <map>
#define LL long long
#define ULL unsigned long long
#define INF 0x3f3f3f3f
#define MOD 1000000007
#define pi acos(-1)
#define fuck() (cout << "--------" << endl)
const int maxn = 100 + 5;
using namespace std;
int maps[maxn][maxn];//障礙1 重要2
int mapss[maxn][maxn];
int link[maxn];
bool vis[maxn];
pair<int,int> g[maxn][maxn];
int row,col;
void init(){
    row = col = 0;
    memset(vis, false, sizeof(vis));
    memset(link, -1, sizeof(link));
    memset(maps, 0, sizeof(maps));
    memset(mapss, 0, sizeof(mapss));
}
bool dfs(int x){
    for(int i=1; i<=col; i++){
        if(!vis[i] && mapss[x][i]){
            vis[i] = true;
            if(link[i] == -1 || dfs(link[i])){
                link[i] = x;
                return true;
            }
        }
    }
    return false;
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        init();
        int x,y;
        int posx,posy;
        scanf("%d%d",&x,&y);
        int blo,imp;
        scanf("%d",&blo);
        for(int i=0; i<blo; i++){
            scanf("%d%d",&posx,&posy);
            maps[posx][posy] = 1;
        }
        scanf("%d",&imp);
        for(int i=0; i<imp; i++){
            scanf("%d%d",&posx,&posy);
            maps[posx][posy] = 2;
        }
        for(int i=1; i<=x; i++){
            bool flag = true;
            for(int j=1; j<=y; j++){
                if(maps[i][j] == 1){
                    if(flag) row++;
                    g[i][j].first = row;
                    flag = false;
                }
                if(maps[i][j] == 2) flag = true;
            }
        }
        for(int j=1; j<=y; j++){
            bool flag = true;
            for(int i=1; i<=x; i++){
                if(maps[i][j] == 1){
                    if(flag) col++;
                    g[i][j].second = col;
                    flag = false;
                }
                if(maps[i][j] == 2) flag = true;
            }
        }
        for(int i=1; i<=x; i++)
            for(int j=1; j<=y; j++)
                if(maps[i][j] == 1)
                    mapss[g[i][j].first][g[i][j].second] = 1;
        int ans = 0;
        for(int i=1; i<=row; i++){
            memset(vis, false, sizeof(vis));
            if(dfs(i)) ans++;
        }
        printf("%d\n",ans);
     }
}