How Many to Be Happy?

How Many to Be Happy?

时间限制: 1 Sec  内存限制: 128 MB

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

来源/分类

ICPC 2017 Daejeon 

---

最小生成树的MST性质的应用。我们想让某一条边一定是最小生成树中的边，只要找到任意一种点集的分配，使得这条边的两个顶点在不同的分配中且边权是连接这两个分配的所有边中最小的那一个。显然只有边权比它小的边才会影响它是不是在最小生成树中。于是我们可以只在图中保留边权小于当前边权的边，看看是否能找到一种点集的分配。显然当这个边的两个顶点在新图中仍然连通时，我们找不到这种分配，于是就需要砍掉若干边使两顶点不连通，于是题目就转化为了最小割问题。 参考博客：

#include<bits/stdc++.h>
#define INF LLONG_MAX/2
#define N 505
using namespace std;

struct ss
{
    int v,next;
    long long flow;
};
int head[N],now_edge=0,S,T;
ss edg[N*2];

void init()
{
    now_edge=0;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v,long long flow)
{
    edg[now_edge]=(ss){v,head[u],flow};
    head[u]=now_edge++;
    edg[now_edge]=(ss){u,head[v],flow};
    head[v]=now_edge++;
}

int dis[N];

int bfs()
{
    memset(dis,0,sizeof(dis));
    queue<int>q;
    q.push(S);
    dis[S]=1;

    while(!q.empty())
    {
        int now=q.front();
        q.pop();

        for(int i=head[now];i!=-1;i=edg[i].next)
        {
            ss &e=edg[i];
            if(e.flow>0&&dis[e.v]==0)
            {
                dis[e.v]=dis[now]+1;
                q.push(e.v);
            }
        }
    }

    if(dis[T]==0)return 0;
    return 1;
}

int current[N];
long long dfs(int x,long long maxflow)
{
    if(x==T)return maxflow;
    for(int i=current[x];i!=-1;i=edg[i].next)
    {
        current[x]=i;

        ss &e=edg[i];
        if(e.flow>0&&dis[e.v]==dis[x]+1)
        {
            long long flow=dfs(e.v,min(maxflow,e.flow));

            if(flow!=0)
            {
                e.flow-=flow;
                edg[i^1].flow+=flow;
                return flow;
            }
        }
    }
    return 0;
}

long long dinic()
{
    long long ans=0,flow;

    while(bfs())
    {
        for(int i=0;i<N;i++)current[i]=head[i];
        while(flow=dfs(S,INF))ans+=flow;
    }
    return ans;
}

int from[N],to[N],w[N];

int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        scanf("%d %d %d",&from[i],&to[i],&w[i]);
    }

    int ans=0;
    for(int i=1;i<=m;i++)
    {
        init();
        for(int j=1;j<=m;j++)
        if(w[j]<w[i])addedge(from[j],to[j],1);

        S=from[i];
        T=to[i];
        ans+=dinic();
    }
    printf("%d\n",ans);
    return 0;
}

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