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How Many to Be Happy?

時間限制: 1 Sec 內存限制: 128 MB

題目描述

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.

輸入

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.

輸出

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.

樣例輸入

3 3
1 2 1
3 1 2
3 2 3

樣例輸出

1

來源/分類

ICPC 2017 Daejeon

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最小生成樹的MST性質的應用。我們想讓某一條邊一定是最小生成樹中的邊，只要找到任意一種點集的分配，使得這條邊的兩個頂點在不同的分配中且邊權是連接這兩個分配的所有邊中最小的那一個。顯然只有邊權比它小的邊才會影響它是不是在最小生成樹中。於是我們可以只在圖中保留邊權小於當前邊權的邊，看看是否能找到一種點集的分配。顯然當這個邊的兩個頂點在新圖中仍然連通時，我們找不到這種分配，於是就需要砍掉若幹邊使兩頂點不連通，於是題目就轉化為了最小割問題。 參考博客：https://blog.csdn.net/shagua_nan/article/details/50936441

#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;
}

View Code

How Many to Be Happy?