HDU 2112——HDU Today

最小路徑的題目，地名的處理是把各個地名標記序號即可。

需要注意的是這題是的公交起點站可到終點站，終點站也可到起點站。另外給出兩個相同地點時，輸出是0.

#include<iostream>#include<cstdio>#include<cstring>using namespace std;#define INF 1000000int map[160][160],dis[160];bool que[160];char place[160][35];int main(){int i,j,sum,s,e,flag,cost;int n,min,k;char ch[35];while(scanf("%d",&n)!=EOF&&n!=-1){memset(que,0,sizeof(que));for(i=1;i<=150;i++)for(j=1;j<=150;j++)map[i][j]=INF;for(i=1;i<=150;i++)map[i][i]=0;scanf("%s%s%*c",place[1],place[2]);sum=2;for(i=1;i<=n;i++){flag=1;scanf("%s%*c",ch);for(j=1;j<=sum;j++)if(strcmp(place[j],ch)==0){flag=0;s=j;break;}if(flag){sum++;strcpy(place[sum],ch);s=sum;}flag=1;scanf("%s%*c",ch);for(j=1;j<=sum;j++)if(strcmp(place[j],ch)==0){flag=0;e=j;break;}if(flag){sum++;strcpy(place[sum],ch);e=sum;}scanf("%d",&cost);if(map[s][e]>cost)map[s][e]=map[e][s]=cost;}que[1]=1;for(i=1;i<=sum;i++)dis[i]=map[1][i];dis[1]=0;for(i=1;i<=sum;i++){min=INF;for(j=1;j<=sum;j++)if(!que[j]&&min>=dis[j]){min=dis[j];k=j;}que[k]=1;for(j=1;j<=sum;j++)if(!que[j]&&dis[j]>dis[k]+map[k][j])dis[j]=dis[k]+map[k][j];}if(strcmp(place[1],place[2])==0){printf("0\n");continue;}if(dis[2]==INF)printf("-1\n");elseprintf("%d\n",dis[2]);}return 0;}