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 }