CF1082G Petya and Graph

Posted 2021-02-03 olinr

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定义(图权 = 图中边权总和 ? 图中点权总和)（空图的图权 =0），求 (n) 个点 (m) 条边的无向图最大权子图。

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第一行为n，m

第二行为点权

接下来为边

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输出一行为最大权子图

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4 5
1 5 2 2
1 3 4
1 4 4
3 4 5
3 2 2
4 2 2
   
    
3 3
9 7 8
1 2 1
2 3 2
1 3 3

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8
    
   
0

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(1leq n leq 10^3, m leq 10 ^3)

(color{#0066ff}{ 题解 })

最大流最小割定理

S向所有边建容量为边权的边

所有点向T建容量为点权的边

每条边向对应的两个点建容量为inf的边

(ans=sum 边权-最大流（最小割）)

#include<bits/stdc++.h>
#define LL long long
LL in() {
    char ch; LL x = 0, f = 1;
    while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
    for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
    return x * f;
}
const int maxn = 1e4 + 10;
const int inf = 0x7fffffff;
struct node {
    int to;
    LL dis;
    node *nxt, *rev;
    node(int to = 0, LL dis = 0, node *nxt = NULL):to(to), dis(dis), nxt(nxt) {}
    void *operator new(size_t) {
        static node *S = NULL, *T = NULL;
        return (S == T) && (T = (S = new node[1024]) + 1024), S++;
    }
};
int n, m, s, t;
int dep[maxn];
std::queue<int> q;
node *head[maxn], *cur[maxn];
void add(int from, int to, LL dis) {
    head[from] = new node(to, dis, head[from]);
}
void link(int from, int to, int dis) {
    add(from, to, dis), add(to, from, 0);
    head[from]->rev = head[to];
    head[to]->rev = head[from];
}
bool bfs() {
    for(int i = s; i <= t; i++) dep[i] = 0, cur[i] = head[i];
    q.push(s);
    dep[s] = 1;
    while(!q.empty()) {
        int tp = q.front(); q.pop();
        for(node *i = head[tp]; i; i = i->nxt)
            if(!dep[i->to] && i->dis) 
                dep[i->to] = dep[tp] + 1, q.push(i->to);
    }
    return dep[t];
}
LL dfs(int x, LL change) {
    if(x == t || !change) return change;
    LL flow = 0, ls;
    for(node *i = cur[x]; i; i = i->nxt) {
        cur[x] = i;
        if(dep[i->to] == dep[x] + 1 && (ls = dfs(i->to, std::min(change, i->dis)))) {
            flow += ls;
            change -= ls;
            i->dis -= ls;
            i->rev->dis += ls;
            if(!change) break;
        }
    }
    return flow;
}
LL dinic() {
    LL flow = 0;
    while(bfs()) flow += dfs(s, inf);
    return flow;
}
LL ans;
int main() {
    n = in(), m = in();
    s = 0, t = n + m + 1;
    LL x, y, z;
    for(int i = 1; i <= n; i++) z = in(), link(m + i, t, z);
    for(int i = 1; i <= m; i++) {
        x = in(), y = in(), z = in();
        link(s, i, z);
        link(i, m + x, inf);
        link(i, m + y, inf);
        ans += z;
    }
    printf("%lld
", ans - dinic());
    return 0;
}