poj 1603 Risk_spfa向前星

时间:2024-08-12 16:34:38

poj终于到100题,贴个代码纪念一下,hdu 到400题再贴

题意:有20个城市,接下来有19行告诉你,i城市与n个城市相连,图是双向的,然后叫你求x到y的最小经过几个城市

#include <iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
#define N 210
#define INF 0xffffff
int num;
struct node{
int u,v,next,d;
}edge[N],tedge[30];
int head[N],flag[N],dist[N];
void add(int u,int v,int d){
edge[num].v=v;
edge[num].d=d;
edge[num].next=head[u];
head[u]=num++;
}
int spfa(int x,int y){
int i,sum=0;
for(i=1;i<=20;i++){
dist[i]=INF;
flag[i]=0;
}
dist[x]=0;
queue<int>q;
q.push(x);
while(!q.empty()){
int tmp=q.front();
q.pop();
flag[tmp]=0;
for(i=head[tmp];i!=-1;i=edge[i].next){
int v=edge[i].v;
int d=edge[i].d;
if(dist[tmp]+d<dist[v]){
dist[v]=dist[tmp]+d;
if(!flag[v]){
q.push(v);
flag[v]=1;
}
}
}
}
return dist[y];
}
int main(int argc, char** argv) {
int Case=1,n,i,x,j,y,ans;
while(scanf("%d",&n)!=EOF){
printf("Test Set #%d\n",Case++);
num=0;
memset(head,-1,sizeof(head));
for(i=0;i<n;i++){
scanf("%d",&x);
add(1,x,1);
add(x,1,1);
}
for(i=2;i<=19;i++){
scanf("%d",&n);
for(j=0;j<n;j++){
scanf("%d",&x);
add(i,x,1);
add(x,i,1);
}
}
scanf("%d",&n);
for(i=0;i<n;i++){
scanf("%d%d",&x,&y);
ans=spfa(x,y);
printf("%d to %d: %d\n",x,y,ans);
}
printf("\n");
}
return 0;
}