码迷,mamicode.com
首页 > 其他好文 > 详细

CF1082G Petya and Graph

时间:2019-01-14 10:53:04      阅读:203      评论:0      收藏:0      [点我收藏+]

标签:最大流最小割   graph   main   ++   接下来   dep   return   i++   ==   

\(\color{#0066ff}{ 题目描述 }\)

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。

\(\color{#0066ff}{输入格式}\)

第一行为n,m

第二行为点权

接下来为边

\(\color{#0066ff}{输出格式}\)

输出一行为最大权子图

\(\color{#0066ff}{输入样例}\)

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

\(\color{#0066ff}{输出样例}\)

8
    
   
0

\(\color{#0066ff}{数据范围与提示}\)

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

\(\color{#0066ff}{ 题解 }\)

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\n", ans - dinic());
    return 0;
}

CF1082G Petya and Graph

标签:最大流最小割   graph   main   ++   接下来   dep   return   i++   ==   

原文地址:https://www.cnblogs.com/olinr/p/10265034.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!