HDU1595

SPFA

题意：给定一张图，从1到n。

求1到n的最短路，但是可能存在某条路被阻断了，即为inf了。

这条路必定是存在最短路中。所以枚举最短路中的路径。。。。

View Code 
  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<stdlib.h>
  4 #include<queue>
  5 #include<algorithm>
  6 using namespace std;
  7 const int maxn = 1005;
  8 const int maxm = 1005*500;
  9 const int inf = 9999999;
 10 struct node{
 11     int u,val,next;
 12 }edge[ maxm*2 ];
 13 int head[ maxn ],cnt;
 14 int dis[ maxn ],vis[ maxn ],path[ maxn ];
 15 int n,m,CNT;
 16 struct node22{
 17     int x,y;
 18 }road[ maxn ];
 19 int mat[ maxn ][ maxn ];
 20 void init(){
 21     cnt=0;
 22     CNT=0;
 23     memset( path,-1,sizeof(path) );
 24     memset( head,-1,sizeof( head ));
 25 }
 26 void addedge( int a,int b,int c ){
 27     edge[ cnt ].u=b;
 28     edge[ cnt ].val=c;
 29     edge[ cnt ].next=head[ a ];
 30     head[ a ]=cnt++;
 31 }
 32 void changeit(const int &x, const int &y, int va)  
 33 {  
 34     for( int i=head[x];i!=-1;i=edge[i].next ){
 35         if( edge[i].u==y ){
 36             edge[i].val=va;
 37         }
 38     }
 39 }  
 40 
 41 void spfa( int s ){
 42     for( int i=1;i<=n;i++ ){
 43         dis[i]=inf;
 44         vis[i]=0;
 45         //path[i]=s;
 46     }
 47     queue<int>q;
 48     while( !q.empty() )
 49         q.pop();
 50     vis[s]=1;
 51     dis[s]=0;
 52     q.push( 1 );
 53     while( !q.empty() ){
 54         int now=q.front();
 55         q.pop();
 56         vis[ now ]=0;
 57         for( int i=head[ now ];i!=-1;i=edge[i].next ){
 58             int next=edge[i].u;
 59             if( dis[next]>dis[now]+edge[i].val ){
 60                 dis[next]=dis[now]+edge[i].val;
 61                 path[next]=now;
 62                 if( vis[next]==0 ){
 63                     vis[next]=1;
 64                     q.push(next);
 65                 }
 66             }
 67         }
 68     }
 69     return ;
 70 }
 71 
 72 int main(){
 73     while( scanf("%d%d",&n,&m)!=EOF ){
 74         int a,b,c;
 75         init();
 76         while( m-- ){
 77             scanf("%d%d%d",&a,&b,&c);
 78             addedge( a,b,c );
 79             addedge( b,a,c );
 80             mat[a][b]=mat[b][a]=c;
 81         }
 82         spfa( 1 );
 83         for( int i=n;i!=1;i=path[i] ){
 84             road[ CNT ].x=i;
 85             road[ CNT ].y=path[i];
 86             CNT++;
 87         }
 88         int ans=0;
 89         for(int i = 0; i < CNT; i++){  
 90             changeit(road[i].x, road[i].y, inf);  
 91             changeit(road[i].y, road[i].x, inf);   
 92             spfa(1);  
 93             if(dis[n] > ans&&dis[n]!=inf) ans = dis[n];  
 94             changeit(road[i].x, road[i].y, mat[road[i].x][road[i].y]);  
 95             changeit(road[i].y, road[i].x, mat[road[i].x][road[i].y]);  
 96         }  
 97         printf("%d\n", ans);
 98     }
 99     return 0;
100 }

 

posted @ 2013-02-21 23:26  xxx0624  阅读(412)  评论(0)  编辑  收藏  举报