「TJOI2018」智力竞赛

题解好难看啊。。。
就是求可重路径覆盖之后最大化剩余点的最小权值
二分答案后就是一个可重复路径覆盖
处理出可达点做二分图匹配就好了

#include<cstdio>
#include<cstring>
#include<algorithm>
#define gc getchar()
#define pc putchar
inline int read() {
    int x = 0,f = 1;
    char c = getchar();
    while(c < '0' || c > '9') c = gc;
    while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc;
    return x * f;
}
void print(int x) {
    if(x < 0) {
        pc('-');
        x = -x;
    }
    if(x >= 10) print(x / 10);
    pc(x % 10 + '0');
 }
const int maxn = 507;
int n,m;
bool mp[maxn][maxn];
int val[maxn];
int a[maxn];
void floyd() {
    for(int k = 1;k <= n;++ k)
        for(int i = 1;i <= n;++ i)
            for(int j = 1;j <= n;++ j)
                mp[i][j] |= mp[i][k] & mp[k][j];
}
int vis[maxn];
int tot = 0;
int bel[maxn];
bool find(int x,int fa) 
{
    for(int i = 1;i <= tot;++ i) 
	{
        if(vis[i] != fa && mp[a[x]][a[i]]) 
		{
            vis[i] = fa;
            if(!bel[i] || find(bel[i],fa)) 
			{
                bel[i] = x;
                return true;
            }
        }
    }
    return false;
}
int check(int x) {
    tot = 0;
    for(int i = 1;i <= m;++ i) 
	     if(val[i] < x)  //统计有效点有多少个 
		    a[++ tot] = i;
    int ret = tot;
    memset(bel,0,sizeof bel);
    for(int i = 1;i <= tot;++ i) 
	{
        if(find(i,i)) 
		   ret --;
    }
    return ret; //返回最小路径覆盖的值 
}
int main() {
    //freopen("contest2.in","r",stdin);
    n = read() + 1, m = read();
    int mx = 0;
    for(int k,i = 1;i <= m;++ i) 
	{
        val[i] = read();
        mx = std::max(mx,val[i]);
        k = read();
        for(int v,j = 1;j <= k;++ j) 
		{
                v = read();
                mp[i][v] = 1;
        } 
    }
    floyd();
    int ans = -1;
    int l = 1,r = mx;
    while(l <= r) 
	{
        int mid = l + r >> 1;
        if(check(mid) <= n)  //如果覆盖成功的话，则左边界可以再增加 
		    l = mid + 1,ans = mid;
        else r = mid - 1;
    }
    if(l <= mx) 
    	print(ans),pc('\n');
    else 
	    puts("AK");
    return 0;
}