標籤：最短時間 mono problem 審核 神經網路 line 連結 directory ons
最短路Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 42585 Accepted Submission(s): 18653
所以如今他們想要尋找最短的從商店到賽場的路線，你能夠協助他們嗎？
N=M=0表示輸入結束。
接下來M行，每行包含3個整數A。B，C（1<=A,B<=N,1<=C<=1000）,表示在路口A與路口B之間有一條路。我們的工作人員須要C分鐘的時間走過這條路。
輸入保證至少存在1條商店到賽場的路線。
2 11 2 33 31 2 52 3 53 1 20 0
32
AC-code:
Dijkstra：
#include<cstdio>#define max 0x3f3f3f3f#define min(a,b) (a>b?b:a)int n,dis[105],cost[105][105];void dijkstra(int a){int i,vis[105];for(i=1;i<=n;i++){dis[i]=max;vis[i]=0;}dis[a]=0;while(1){int v=-1;for(i=1;i<=n;i++)if(!vis[i]&&(v==-1||dis[i]<dis[v]))v=i;if(v==-1)break;vis[v]=1;for(i=1;i<=n;i++)dis[i]=min(dis[i],dis[v]+cost[v][i]);}}int main(){int m,i,j,a,b,c;while(scanf("%d%d",&n,&m),n||m){for(i=1;i<=n;i++)for(j=1;j<=n;j++)cost[i][j]=max;for(i=1;i<=m;i++){scanf("%d%d%d",&a,&b,&c);cost[a][b]=cost[b][a]=min(cost[a][b],c);}dijkstra(1);printf("%d\n",dis[n]);}return 0;}
SPFA：
#include<cstdio>#include<queue>#include<cstring>#define INF 0x3f3f3f3fint dis[105],vis[105],n,head[105*105],num;using namespace std;struct node{int from,to,val,next;}A[105*105];void spfa(int sx){queue<int>q;memset(dis,INF,sizeof(dis));memset(vis,0,sizeof(vis));q.push(sx);dis[sx]=0;vis[sx]=1;while(!q.empty()){int u=q.front();q.pop();vis[u]=0;for(int i=head[u];i!=-1;i=A[i].next){int v=A[i].to;if(dis[v]>dis[u]+A[i].val){dis[v]=dis[u]+A[i].val;if(!vis[v]){vis[v]=1;q.push(v);}}}}printf("%d\n",dis[n]);}void chan(int a,int b,int c){node B={a,b,c,head[a]};A[num]=B;head[a]=num++;}int main(){int i,m,a,b,c;while(scanf("%d%d",&n,&m),n||m){num=0;memset(head,-1,sizeof(head));for(i=0;i<m;i++){scanf("%d%d%d",&a,&b,&c);chan(a,b,c);chan(b,a,c);}spfa(1);}return 0;}
q=dijkstra&t=blog" target="_blank">dijkstra /
q=HDU&t=blog" target="_blank">HDU /
q=圖論&t=blog" target="_blank">圖論 /
q=SPFA&t=blog" target="_blank">SPFA /
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