TJOI2018 智力竞赛

Posted 2021-01-26 akoasm

这题做的真的让我没了智力...

题目大意：
从n个点的图中选出n + 1条链问是否能够全部覆盖（可相交），如果不能，输出覆盖的最小点权最小值最大是多少？

思路：
最小值最大问题考虑二分，二分答案用二分图匹配去求链覆盖即可，记得floyd处理闭包。

吐槽：
调了半天发现是二分上界太大...wtf？

Code:

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 510;
const int INF = 1e9;
int v[MAXN];
int f[MAXN][MAXN];
int match[MAXN];
int a[MAXN];
int l;
int m;
int cnt;
int n;
int ans;
int mx;
int vis[MAXN];
int mk[MAXN][MAXN];
int r;
int read () {
    int q=0,f=1;char ch=getchar();
    while(!isdigit(ch)){
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(isdigit(ch)){
        q=q*10+ch-'0';ch=getchar();
    }
    return q*f;
}
int dfs(int x) {
    for(int i = 1;i <= cnt; ++i) {
        if(!vis[i] and mk[a[x]][a[i]]) {
            vis[i] = 1;
            if(!match[i] or dfs(match[i])) {
                match[i] = x;
                return true;
            }
        }
    }
    return false;
}

int ok(int mid) {
    cnt = 0;
    for(int i = 1;i <= m; ++i) {
        if(v[i] < mid) a[++cnt] = i;
    }
    int res = cnt;
    for(int i = 1;i <= cnt; ++i) {
        memset(vis,0,sizeof vis);
        res -= dfs(i);
    }
    return res;
}
void floyd() {
    for(int k = 1;k <= m; ++k) {
        for(int i = 1;i <= m; ++i) {
            for(int j = 1;j <= m; ++j) {
                mk[i][j] |= mk[i][k] & mk[k][j];
            }
        }
    }
}
int main () {
    n = read(),m = read();
     ++ n;
    for(int i = 1;i <= m; ++i) {
        v[i] = read();int tmp = read();
        mx = max(mx,v[i]);
        for(int j = 1;j <= tmp; ++j) {
            int x = read();
            mk[i][x] = 1;
        }
    }
    floyd();
    l = 1;
    r = mx;
    ans = 0;
    while(l <= r) {
        int mid = (l + r) >> 1;
        memset(match,0,sizeof match);
        if(ok(mid) <= n) {
            l = mid + 1;
            ans = mid;
        }
        else r = mid - 1;
    }
    if(ans >= mx) puts("AK");
    else printf("%d
",ans);
    return 0;
}