hdu 2433 Travel（还不会）

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

#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<stdio.h>
#include<numeric>//STL数值算法头文件
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板类头文件
using namespace std;
#define N 110
#define M 6020
const int INF=0x3f3f3f3f;
const int maxn=1010;

struct Edgs
{
    int to,next;
} E[M];

struct Node
{
    int x, y;
} node[M];

int d[N], pre[N][N], num[N][N], sum[N], head[N];
int cnt, n, m;
bool flag = true;
bool vis[N];

void add_edgs(int u, int v)
{
    E[cnt].to = v;
    E[cnt].next = head[u];
    head[u] = cnt++;
}

void init()
{
    memset(num, 0, sizeof(num));
    memset(head, -1, sizeof(head));
    cnt = 0;
    int x, y;
    for (int i = 0; i < m; i++)
    {
        scanf("%d%d", &x, &y);
        node[i].x = x;
        node[i].y = y;
        num[x][y]++;
        num[y][x]++;
        add_edgs(x, y);
        add_edgs(y, x);
    }
}

void bfs(int s)
{
    queue<int> q;
    for (int i = 1; i <= n; i++)
    {
        d[i] = INF;
        vis[i] = false;
    }
    d[s] = 0;
    pre[s][s] = 0;
    vis[s] = 1;
    q.push(s);
    while (!q.empty())
    {
        int u = q.front();
        q.pop();

        for (int i = head[u]; i != -1; i = E[i].next)
        {
            int v = E[i].to;
            if (!vis[v])
            {
                d[v] = d[u] + 1;
                vis[v] = 1;
                pre[s][v] = u;
                q.push(v);
            }
        }
    }
    sum[s] = 0;
    for (int i = 1; i <= n; i++)
    {
        if (d[i] == INF)
        {
            flag = false;
            return ;
        }
        sum[s] += d[i];
    }
}

int bfs2(int s)
{
    queue<int> q;
    for (int i = 1; i <= n; i++)
    {
        vis[i] = false;
        d[i] = INF;
    }
    d[s] = 0;
    vis[s] = true;
    q.push(s);

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

        for (int i = head[u]; i != -1; i = E[i].next)
        {
            int v = E[i].to;
            if (num[u][v] && !vis[v])
            {
                d[v] = d[u] + 1;
                vis[v] = true;
                q.push(v);
            }
        }
    }

    int ans = 0;
    for (int i = 1; i <= n; i++)
    {
        if (d[i] == INF)
            return -1;
        ans += d[i];
    }
    return ans;
}

void solve()
{

    flag = true;
    for (int i = 1; i <= n; i++)
    {
        if (flag)
            bfs(i);
        else
            break;
    }

    for (int i = 0; i < m; i++)
    {

        if (!flag)
        {
            printf("INF\n");
            continue;
        }
        int x = node[i].x;
        int y = node[i].y;

        int ans = 0, j;
        for (j = 1; j <= n; j++)
        {
            if (pre[j][x] != y && pre[j][y] != x)
            {
                ans += sum[j];
                continue;
            }
            num[x][y]--;
            num[y][x]--;

            int t = bfs2(j);

            num[y][x]++;
            num[x][y]++;

            if (t == -1)
            {
                printf("INF\n");
                break;
            }
            ans += t;
        }
        if (j == n + 1)
            printf("%d\n", ans);
    }
}

int main()
{
    while (scanf("%d%d", &n, &m) != EOF)
    {
        init();
        solve();
    }
    return 0;
}