Codeforces 1082 G

G - Petya and Graph

思路:

最大权闭合子图

对于每条边,如果它选了,那么它连的的两个点也要选

边权为正,点权为负,那么就是求最大权闭合子图

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const LL INF = 1LL<<40;
const int N = 2e3 + 100;
int level[N], iter[N];
struct edge {
    int to;
    LL w;
    int rev;
};
vector<edge>g[N];
void add_edge(int u, int v, LL w) {
    g[u].pb(edge{v, w, g[v].size()});
    g[v].pb(edge{u, 0, g[u].size()-1});
}
void bfs(int s) {
    mem(level, -1);
    queue<int>q;
    level[s] = 0;
    q.push(s);
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        for (int i = 0; i < g[u].size(); i++) {
            edge e = g[u][i];
            if(e.w > 0 && level[e.to] < 0) {
                level[e.to] = level[u] + 1;
                q.push(e.to);
            }
        }
    }
}
LL dfs(int u, int t, LL f) {
    if(u == t ) return f;
    for (int &i = iter[u]; i < g[u].size(); i++) {
        edge &e = g[u][i];
        if(e.w > 0 && level[u] < level[e.to]) {
            LL d = dfs(e.to, t, min(f, e.w));
            if(d > 0) {
                e.w -= d;
                g[e.to][e.rev].w +=d;
                return d;
            }
        }
    }
    return 0;
}
LL max_flow(int s, int t) {
    LL flow = 0;
    while(true) {
        bfs(s);
        if(level[t] < 0) return flow;
        LL f;
        mem(iter, 0);
        while ((f = dfs(s, t, INF)) > 0) {
            flow += f;
        }
    }
}
int main() {
    int n, m, w, u, v;
    scanf("%d %d", &n, &m);
    int s = 0, t = n+m+1;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &w);
        add_edge(i, t, w);
    }
    LL sum = 0;
    for (int i = 1; i <= m; i++) {
        scanf("%d %d %d", &u, &v, &w);
        sum += w;
        add_edge(i+n, u, INF);
        add_edge(i+n, v, INF);
        add_edge(s, i+n, w);
    }
    printf("%lld
", sum - max_flow(s, t));
    return 0;
}
原文地址:https://www.cnblogs.com/widsom/p/10040337.html