Codeforces 1082 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<<;
const int N = 2e3 + ;
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, , g[u].size()-});
}
void bfs(int s) {
mem(level, -);
queue<int>q;
level[s] = ;
q.push(s);
while (!q.empty()) {
int u = q.front();
q.pop();
for (int i = ; i < g[u].size(); i++) {
edge e = g[u][i];
if(e.w > && level[e.to] < ) {
level[e.to] = level[u] + ;
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 > && level[u] < level[e.to]) {
LL d = dfs(e.to, t, min(f, e.w));
if(d > ) {
e.w -= d;
g[e.to][e.rev].w +=d;
return d;
}
}
}
return ;
}
LL max_flow(int s, int t) {
LL flow = ;
while(true) {
bfs(s);
if(level[t] < ) return flow;
LL f;
mem(iter, );
while ((f = dfs(s, t, INF)) > ) {
flow += f;
}
}
}
int main() {
int n, m, w, u, v;
scanf("%d %d", &n, &m);
int s = , t = n+m+;
for (int i = ; i <= n; i++) {
scanf("%d", &w);
add_edge(i, t, w);
}
LL sum = ;
for (int i = ; 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\n", sum - max_flow(s, t));
return ;
}
Codeforces 1082 G - Petya and Graph的更多相关文章
- CodeForces 1082 G Petya and Graph 最大权闭合子图。
题目传送门 题意:现在有一个图,选择一条边,会把边的2个顶点也选起来,最后会的到一个边的集合 和一个点的集合 , 求边的集合 - 点的集合最大是多少. 题解:裸的最大权闭合子图. 代码: #inclu ...
- G. Petya and Graph(经典项目与项目消耗问题)(网络流)
题:https://codeforces.com/contest/1082/problem/G 题意:给定有边权和点权的图,问你选一些边,然sum边-sum点最大(点权被多次用为公共点只会减一次) 分 ...
- CF1082G:G. Petya and Graph(裸的最大闭合权图)
Petya has a simple graph (that is, a graph without loops or multiple edges) consisting of n n vertic ...
- CodeForces 1082 D Maximum Diameter Graph
题目传送门 题意:现在有n个点,每个点的度数最大为di,现在要求你构成一棵树,求直径最长. 题解:把所有度数为2的点先扣出来,这些就是这颗树的主干,也就是最长的距离. 然后我们把度数为2的点连起来,之 ...
- Petya and Graph/最大权闭合子图、最小割
原题地址:https://codeforces.com/contest/1082/problem/G G. Petya and Graph time limit per test 2 seconds ...
- Petya and Graph(最小割,最大权闭合子图)
Petya and Graph http://codeforces.com/contest/1082/problem/G time limit per test 2 seconds memory li ...
- [codeforces 549]G. Happy Line
[codeforces 549]G. Happy Line 试题描述 Do you like summer? Residents of Berland do. They especially love ...
- CodeForces 794 G.Replace All
CodeForces 794 G.Replace All 解题思路 首先如果字符串 \(A, B\) 没有匹配,那么二元组 \((S, T)\) 合法的一个必要条件是存在正整数对 \((x,y)\), ...
- NEERC 2016-2017 Probelm G. Game on Graph
title: NEERC 2016-2017 Probelm G. Game on Graph data: 2018-3-3 22:25:40 tags: 博弈论 with draw 拓扑排序 cat ...
随机推荐
- Bootstrap表单样式
<form class="form-horizontal" role="form"> <fieldset> <legend> ...
- 浅谈k短路算法
An Old but Classic Problem 给定一个$n$个点,$m$条边的带正权有向图.给定$s$和$t$,询问$s$到$t$的所有权和为正路径中,第$k$短的长度. Notice 定义两 ...
- Codeforces 841D Leha and another game about graph - 差分
Leha plays a computer game, where is on each level is given a connected graph with n vertices and m ...
- uniGUI试用笔记(十)
今天用LoadRunner对uniGUI的Standalone模式的程序进行了一次压力测试,程序采用三层模式,将应用服务器与Web服务器分离,由于条件限制,数据库.应用服务和Web服务都部署在同一条云 ...
- Ubuntu 汉字显示一半的修复
部分汉字(例如“复”)显示不正常,只显示一半,其原因是Ubuntu采用的字体Noto Sans CJK优先显示日文汉字,这一问题可以通过修改配置文件/etc/fonts/conf.avail/64-l ...
- 用vim + xdebug 来追踪thinkphp的执行过程
tree命令的使用几个有实际应用的参数 -a 这是默认的 -d: 只显式目录, 不需要显式目录下的文件 -L: 列出显式的深度. 当前目录下的所有东西为第一级... 在tp下, 有多个Common但是 ...
- FJUT 聪明的商人(树上倍增)题解
思路:求树上两点的距离,显然是dep[u] + dep[v] - 2 * dep[lca],用树上倍增去写. 参考:树上倍增的写法和应用(详细讲解,新手秒懂) 代码: #include<set& ...
- 360搜索引擎取真实地址-python代码
还是个比较简单的,不像百度有加密算法 分析 http://www.so.com/link?url=http%3A%2F%2Fedu.sd.chinamobile.com%2Findex%2Fnews. ...
- Match function in R
Examples: print(match(5, c(1,2,9,5,3,6,7,4,5)))[1] 4 5 %in% c(1,2,9,5,3,6,7,4,5)[1] TRUE ...
- (zhuan) Speech and Natural Language Processing
Speech and Natural Language Processing obtain from this link: https://github.com/edobashira/speech-l ...