hdu 2433 Travel 最短路 dijkstra算法。

Travel

Time Limit: 10000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1819    Accepted Submission(s): 614

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

 

本来不打算写这篇博客的代码的。

因为我的代码根本就是错的，

我是照这篇博客写的：http://www.cppblog.com/luxiuyuan/archive/2012/03/09/167494.aspx

我在看这篇博客的时候，，就对博主的算法感到怀疑。但是他的代码可以A掉，

后来看了discuss

有一组数据：

4 4 1 22 33 4 2 4答案：INF201820

很多代码都是过不了遮住测试数据的不掉的。

对此，我感觉务必绝望啊~~

代码：

#include <stdio.h>#include <string.h>#define MAX 150#define INF 1000000000struct Point{int x, y ;}p[MAX*30] ;int c[MAX][MAX];int graph[MAX][MAX] , sum[MAX];int dis[MAX] ;bool visited[MAX] ;int dijkstra(int n,int s){memset(visited,false,sizeof(visited)) ;for(int i = 1 ; i <= n ; ++i){dis[i] = graph[s][i] ; }dis[s] = 0 ;visited[s] = true ;int sumt = 0 ;for(int i = 1 ; i < n ; ++i){int index = -1 , min = INF ;for(int j = 1 ; j <= n ; ++j){if(!visited[j] && min>dis[j]){index = j ;min = dis[j] ;}}if(index == -1){return INF ;}sumt += min ;visited[index] = true ;for(int j = 1 ; j <= n ; ++j){if(!visited[j] && dis[j]>dis[index]+graph[index][j]){dis[j] = dis[index] + graph[index][j] ;}}}return sumt ;}int main(){int n , m ;while(~scanf("%d%d",&n,&m)){memset(c,0,sizeof(c)) ;for(int i = 0 ; i < MAX ; ++i){for(int j = 0 ; j < MAX ; ++j){graph[i][j] = INF ;}}for(int i = 0 ; i < m ; ++i){int x , y ;scanf("%d%d",&x,&y) ; p[i].x = x , p[i].y = y ;c[x][y]++ ,c[y][x]++ ;graph[x][y] = graph[y][x] = 1 ;}bool flag = true ;int ans = 0 ;for(int i = 1 ; i <= n ; ++i){if(flag){sum[i] = dijkstra(n,i) ;if(sum[i] == INF){flag = false ;}ans += sum[i] ;}}for(int i = 0 ; i < m ; ++i){if(!flag){puts("INF") ;}else{if(c[p[i].x][p[i].y]>1){printf("%d\n",ans) ;}else{graph[p[i].x][p[i].y] = graph[p[i].y][p[i].x] = INF ;int sumx = dijkstra(n,p[i].x) ;int sumy = dijkstra(n,p[i].y) ;if(sumx == INF || sumy == INF){puts("INF") ;}else{printf("%d\n",ans+sumx+sumy-sum[p[i].x]-sum[p[i].y]) ;}graph[p[i].x][p[i].y] = graph[p[i].y][p[i].x] = 1 ;}}}}return 0 ;}

与君共勉