loj6045 「雅礼集训 2017 Day8」价

Posted 2020-09-22 逢山开路 遇水架桥

传送门：https://loj.ac/problem/6045

【题解】

由于存在完美匹配，所以选择k个药就要选择>=k个药材，我们要求的是选择k个药正好选择k个药材。

那么定义选一种减肥药的代价为-pi+INF，选一种药材的代价为INF，这样最小割肯定是恰好选k个

那么 最后答案就是最小割 - Σ(-pi+INF) 【由于减肥药是负的所以要反过来。。。】

中间连的边要设成比inf大的。。

# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int N = 600 + 10, M = 5e5 + 10;
const int mod = 1e9+7, inf = 5e6, INF = 1e9;

# define RG register
# define ST static

int n, m[N], a[N][N], p[N], S, T;
int head[N], nxt[M], to[M], flow[M], tot=1;

inline void add(int u, int v, int fl) {
    ++tot; nxt[tot] = head[u]; head[u] = tot; to[tot] = v; flow[tot] = fl;
}
inline void adde(int u, int v, int fl) {
    add(u, v, fl), add(v, u, 0);
}

namespace MF {
    struct queue {
        int head, tail, q[M];
        inline void set() {
            head = 1, tail = 0;
        }
        inline void push(int x) {
            q[++tail] = x;
        }
        inline void pop() {
            ++head;
        }
        inline int front() {
            return q[head];
        }
        inline bool empty() {
            return head > tail;
        }
    }q;
    
    int c[N], cur[N]; 
    inline bool bfs() {
        for (int i=1; i<=n+n+2; ++i) c[i] = -1;
        q.set(); q.push(S); c[S] = 1;
        while(!q.empty()) {
            int top = q.front(); q.pop();
            for (int i=head[top]; i; i=nxt[i]) {
                if(c[to[i]] != -1 || flow[i] == 0) continue;
                c[to[i]] = c[top] + 1;
                q.push(to[i]);
                if(to[i] == T) return 1;
            }
        }
        return 0;
    }
    
    inline int dfs(int x, int low) {
        if(x == T) return low;
        int r = low, fl;
        for (int i=cur[x]; i; i=nxt[i]) {
            if(c[to[i]] != c[x]+1 || flow[i] == 0) continue;
            fl = dfs(to[i], min(r, flow[i]));
            flow[i] -= fl; flow[i^1] += fl; r -= fl;
            if(flow[i] > 0) cur[x] = i; 
            if(!r) return low;
        }
        if(r == low) c[x] = -1; 
        return low-r;
    }
    
    inline int main() {
        int ans = 0; 
        while(bfs()) {
            for (int i=1; i<=n+n+2; ++i) cur[i] = head[i];
            ans += dfs(S, INF); 
        }
        return ans;
    }
}

int ans = 0;
int main() {
    freopen("z.in", "r", stdin);
    freopen("z.out", "w", stdout); 
    cin >> n;
    for (int i=1; i<=n; ++i) {
        cin >> m[i];
        for (int j=1; j<=m[i]; ++j) {
            scanf("%d", &a[i][j]);
            adde(i, a[i][j] + n, INF);
        }
    }
    for (int i=1; i<=n; ++i) scanf("%d", p+i);
    S = n+n+1, T = n+n+2;
    for (int i=1; i<=n; ++i) adde(i+n, T, inf);
    for (int i=1; i<=n; ++i) adde(S, i, inf-p[i]), ans += inf-p[i];
    ans = MF::main() - ans;
    cout << ans << endl;
    return 0;
}
/*
3 
2 1 2 
2 1 2 
1 3 
-10 20 -3 
*/

View Code