loj#2574. 「TJOI2018」智力竞赛 (路径覆盖)

• 题目链接
• 题解
• 代码

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题目链接

loj#2574. 「TJOI2018」智力竞赛

题解

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

代码

#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) 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; 
} 
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) l = mid + 1,ans = mid; 
        else r = mid - 1; 
    } 
    if(l <= mx) print(ans),pc('\n'); 
    else puts("AK"); 
    return 0; 
}