Travel

Time Limit: 10000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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 4
5 1
1 3
3 2
5 4
2 2
1 2
1 2

Sample Output

INF
INF
INF
INF
2
2

Source

2008 Asia Chengdu Regional Contest Online

题意：每次只能删除一条边，问每个点到所有点的最短路径总和最小。只要任意两个点之间没有路的就输出 INF。

#include<stdio.h>
#include<queue>
using namespace std;
const int N = 105;
const int inf = 99999999;
struct EDG
{
    int u,v,id;
};
vector<EDG>mapt[N];
int dis[N],n,numEdg[N][N],father[N],goEdg[N][N][N],sum[N];

void spfa(int id,int s,bool b)
{
    queue<int>q;
    int inq[N]={0};
    for(int i=1;i<=n;i++)
        dis[i]=inf;
    q.push(s);
    dis[s]=0; father[s]=s;
    while(!q.empty())
    {
        s=q.front(); q.pop();
        inq[s]=0;
        int len=mapt[s].size();
        for(int i=0;i<len;i++)
        if(id!=mapt[s][i].id)
        {
            int v=mapt[s][i].v;
            if(dis[v]>dis[s]+1)
            {
                dis[v]=dis[s]+1;
                if(b)
                 father[v]=s;
                if(inq[v]==0)
                    inq[v]=1,q.push(v);
            }
        }
    }
}
int main()
{
    int m,a,b;
    EDG ss,edg[N*30];
    while(scanf("%d%d",&n,&m)>0)
    {
        for(int i=1;i<=n;i++)
        {
            mapt[i].clear();
            for(int j=1;j<=n;j++)
            {
                numEdg[i][j]=0;
                for(int e=1;e<=n;e++)
                goEdg[i][j][e]=0;
            }
        }

        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&a,&b);
            edg[i].u=a; edg[i].v=b;
            numEdg[a][b]++; numEdg[b][a]++;
            if(numEdg[a][b]>1)
             continue;
            ss.id=i;
            ss.v=b; mapt[a].push_back(ss);
            ss.v=a; mapt[b].push_back(ss);
        }
        int SUM=0;
        for(int i=1;i<=n;i++)
        {
            spfa(-1,i,true);
            sum[i]=0;
            for(int j=1;j<=n;j++)
            if(dis[j]==inf)
            {
                sum[i]=inf; break;
            }
            else sum[i]+=dis[j];
            if(sum[i]==inf)
            {
               SUM=inf; break;
            }
            SUM+=sum[i];
            for(int j=1;j<=n;j++)
            {
                int u,v=j;
                while(father[v]!=v)
                {
                    u=father[v];
                    goEdg[i][u][v]=goEdg[i][v][u]=1;
                    v=u;
                }
            }
        }
        int ans;
        for(int id=1;id<=m;id++)
        {
            if(SUM==inf)
            {
                printf("INF\n"); continue;
            }
            else if(numEdg[edg[id].u][edg[id].v]>1)
            {
                printf("%d\n",SUM); continue;
            }
            ans=0;
            for(int i=1;i<=n;i++)
            if(goEdg[i][edg[id].u][edg[id].v])//最主要的优化
            {
                spfa(id,i,false);
                for(int j=1;j<=n;j++)
                 if(dis[j]==inf)
                {
                    ans=inf; break;
                }
                else ans+=dis[j];
                if(ans==inf)
                    break;
            }
            else ans+=sum[i];
            if(ans==inf)
                printf("INF\n");
            else printf("%d\n",ans);
        }
    }
}