hdu 2433 Travel(枚举删掉的边求取最短路)

Travel

Time Limit: 10000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1872    Accepted Submission(s): 627

Problem Description

      One day, Tom traveled to a country named BGM. BGM is a small country, but there are N (N <= 100) towns in it. Each town products one kind of food, the food will be transported to all the towns. In addition, the trucks will always take the shortest way. There are M (M <= 3000) two-way roads connecting the towns, and the length of the road is 1.
      Let SUM be the total distance of the shortest paths between all pairs of the towns. Please write a program to calculate the new SUM after one of the M roads is destroyed.

 

Input

      The input contains several test cases.
      The first line contains two positive integers N, M. The following M lines each contains two integers u, v, meaning there is a two-way road between town u and v. The roads are numbered from 1 to M according to the order of the input.
      The input will be terminated by EOF.

 

Output

      Output M lines, the i-th line is the new SUM after the i-th road is destroyed. If the towns are not connected after the i-th road is destroyed, please output “INF” in the i-th line.

 

Sample Input

5 45 11 33 25 42 21 21 2

 

Sample Output

INFINFINFINF22

 

Source

2008 Asia Chengdu Regional Contest Online

题目分析：
记录以某一个点为源点的最短路的生成树中的边，然后每次删掉边，判断这条边是否在这个生成树中，如果在重新求取最短路，如果不在，那么当前的最短路生成树没有变化
 

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#include <queue>#define INF (0x3f3f3f3f)#define N 107using namespace std;int n,m;int mp[N][N];int pre[N][N];int dis[N];bool vis[N];int sum[N];struct Edge{    int u,v;}e[30*N];int spfa ( int s , int f ){    memset ( vis , 0 , sizeof ( vis ) );    memset ( dis, 0 , sizeof ( dis ) );    queue<int> q;    q.push ( s );    vis[s] = 1;    while ( !q.empty() )    {        int u = q.front();        q.pop();        for ( int i = 1 ; i <= n ; i++ )            if ( mp[u][i] && !vis[i] )            {                dis[i] = dis[u] + 1;                vis[i] = 1;                if ( f ) pre[s][i] = u;                q.push ( i );            }    }    int sum = 0;    for ( int i = 1 ; i <= n ; i++ )        if ( !dis[i]&& i != s ) return INF;        else sum += dis[i];    return sum;}int main ( ){    while ( ~scanf ( "%d%d" , &n , &m ) )    {        int u , v;        memset ( mp , 0 , sizeof ( mp ) );        memset ( pre , 0 , sizeof ( pre ) );        for ( int i = 0 ; i < m ; i++ )        {            scanf ( "%d%d" , &u , &v );            e[i].u = u;            e[i].v = v;            mp[u][v]++;            mp[v][u]++;        }        int temp = 0;        bool flag = 1;        for ( int i = 1 ; i <= n ; i++ )        {            sum[i] = spfa ( i , 1 );            temp += sum[i];            if ( temp >= INF )            {                flag = 0;                break;            }        }        for ( int i = 0 ; i < m ; i++ )        {            int ans = 0;            u = e[i].u , v = e[i].v;            if ( flag && mp[u][v] == 1 )            {                ans = 0;                for ( int j = 1 ; j <= n ; j++ )                {                    if ( pre[j][u] != v && pre[j][v] != u )                        ans += sum[j];                    else                    {                        mp[u][v]--;                        mp[v][u]--;                        ans += spfa ( j , 0 );                        mp[u][v]++;                        mp[v][u]++;                        if ( ans >= INF )                            break;                    }                }            }            else ans = temp;            if ( ans >= INF ) puts ( "INF" );            else printf ( "%d\n" , ans );        }     }}