E: How Many to Be Happy?
时间限制: 1 Sec 内存限制: 128 MB
提交: 75 解决: 29
[提交] [状态] [讨论版] [命题人:admin]
题目描述
Let G be a connected simple undirected graph where each edge has an associated weight. Let’s consider the popular MST (Minimum Spanning Tree) problem. Today, we will see, for each edge e, how much modification on G is needed to make e part of an MST for G. For an edge e in G, there may already exist an MST for G that includes e. In that case, we say that e is happy in G and we define H(e) to be 0. However, it may happen that there is no MST for G that includes e. In such a case, we say that e is unhappy in G. We may remove a few of the edges in G to make a connected graph G′ in which e is happy. We define H(e) to be the minimum number of edges to remove from G such that e is happy in the resulting graph G′.
Figure E.1. A complete graph with 3 nodes.
Consider the graph in Figure E.1. There are 3 nodes and 3 edges connecting the nodes. One can easily see that the MST for this graph includes the 2 edges with weights 1 and 2, so the 2 edges are happy in the graph. How to make the edge with weight 3 happy? It is obvious that one can remove any one of the two happy edges to achieve that.
Given a connected simple undirected graph G, your task is to compute H(e) for each edge e in G and print the total sum.
输入
Your program is to read from standard input. The first line contains two positive integers n and m, respectively, representing the numbers of vertices and edges of the input graph, where n ≤ 100 and m ≤ 500. It is assumed that the graph G has n vertices that are indexed from 1 to n. It is followed by m lines, each contains 3 positive integers u, v, and w that represent an edge of the input graph between vertex u and vertex v with weight w. The weights are given as integers between 1 and 500, inclusive.
输出
Your program is to write to standard output. The only line should contain an integer S, which is the sum of H(e) where e ranges over all edges in G.
样例输入
3 3 1 2 1 3 1 2 3 2 3
样例输出
1
#include<bits/stdc++.h>
using namespace std;
const int N=105,M=1005;
int n,m,head[N],cnt;
int Next[M],To[M],flow[M];
struct node{
int u,v,w;
}a[M];
void addedge(int u,int v,int f){
Next[cnt]=head[u],To[cnt]=v,flow[cnt]=f,head[u]=cnt++;
}
queue<int>Q;
int level[N];
int bfs(int S,int T){
while(!Q.empty())Q.pop();
memset(level,0,sizeof(level));
level[S]=1;
Q.push(S);
while(!Q.empty()){
int u=Q.front();
for(int i=head[u];~i;i=Next[i]){
int v=To[i];
if(!flow[i]||level[v])continue;
level[v]=level[u]+1;
Q.push(v);
if(v==T)return 1;
}
Q.pop();
}
return 0;
}
int dfs(int u,int minf,int T){
if(u==T)return minf;
int res=0;
for(int i=head[u];~i;i=Next[i]){
int v=To[i];
if(level[v]!=level[u]+1||!flow[i])continue;
int tmp=dfs(v,min(minf,flow[i]),T);
minf-=tmp,res+=tmp;
flow[i]-=tmp,flow[i^1]+=tmp;
if(minf==0)return res;
}
level[u]=0;
return res;
}
int dinic(int S,int T){
for(int i=0;i<cnt;i++)flow[i]=1;
int res=0;
while(bfs(S,T))
res+=dfs(S,0x3f3f3f3f,T);
return res;
}
int main(){
scanf("%d %d",&n,&m);
for(int i=1;i<=m;i++)
scanf("%d%d%d",&a[i].u,&a[i].v,&a[i].w);
sort(a+1,a+1+m,[](node a,node b){return a.w<b.w;});
cnt=0;
memset(head,-1,sizeof(head));
int ans=0,l=1,r=1;
for(;l<=m;l++){
while(a[r].w<a[l].w){
addedge(a[r].u,a[r].v,1);
addedge(a[r].v,a[r].u,1);
r++;
}
ans+=dinic(a[l].u,a[l].v);
}
printf("%d\n",ans);
return 0;
}