POJ 1548 Robots

题意：乍一看还以为是小白上那题dp，事实上不是，就是求一共几个机器人能够覆盖全部路径

思路：最小路径覆盖问题。一个点假设在还有一个点右下方，就建边。然后跑最小路径覆盖就可以

代码：

#include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;const int N = 25 * 25;int x[N], y[N], cnt = 1;vector
         
           g[N];bool judge(int i, int j) {	return x[j] >= x[i] && y[j] >= y[i];}int left[N], vis[N];bool dfs(int u) {	for (int i = 0; i < g[u].size(); i++) {		int v = g[u][i];		if (vis[v]) continue;		vis[v] = 1;		if (left[v] == -1 || dfs(left[v])) {			left[v] = u;			return true;		}	}	return false;}int hungary() {	int ans = 0;	memset(left, -1, sizeof(left));	for (int i = 0; i < cnt; i++) {		memset(vis, 0, sizeof(vis));		if (dfs(i)) ans++;	}	return ans;}int main() {	while (~scanf("%d%d", &x[0], &y[0])) {		if (x[0] == -1 && y[0] == -1) break;		while (~scanf("%d%d", &x[cnt], &y[cnt])) {			if (x[cnt] == 0 && y[cnt] == 0) break;			cnt++;		}		for (int i = 0; i < cnt; i++) {			g[i].clear();			for (int j = 0; j < i; j++) {				if (judge(i, j)) g[i].push_back(j);				if (judge(j, i)) g[j].push_back(i);			}		}		printf("%d\n", cnt - hungary());		cnt = 1;	}	return 0;}