Poj 1603 risk

The shortest path.

The first 19 rows are adjacent, undirected graphs, and the last query is the shortest.

Floyd is the simplest. But I use spfa. In fact, it is to find the most short circuit, but there is no distance, but each time + 1.

Note that the last line needs to output a blank line. Contribute to the release of PE.

#include<cstdio>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<map>#include<stack>#include<iostream>#include<list>#include<set>#include<cmath>#define INF 0x7fffffff#define eps 1e-6#define LL long longusing namespace std;vector<int>g[21];int dis[21][21];bool sp[21];int n,m;void SPFA(int start){    queue<int>q;    bool vis[21];    memset(vis,0,sizeof(vis));    q.push(start);    dis[start][start]=0;    vis[start]=1;    while(!q.empty())    {        int u=q.front();q.pop();        vis[u]=0;        for(int j=0;j<g[u].size();j++)        {            int v=g[u][j];            if(dis[start][v]>dis[start][u]+1)            {                dis[start][v]=dis[start][u]+1;                if(!vis[v])                {                    vis[v]=1;                    q.push(v);                }            }        }    }}int main(){    int nn=1;    while(scanf("%d",&m)!=EOF)    {        for(int i=1;i<21;i++)            g[i].clear(),sp[i]=0;        int v;        while(m--)        {            scanf("%d",&v);            g[1].push_back(v);            g[v].push_back(1);        }        for(int i=2;i<20;i++)        {            scanf("%d",&m);            while(m--)            {                scanf("%d",&v);                g[i].push_back(v);                g[v].push_back(i);            }        }        for(int i=1;i<21;i++)            for(int j=1;j<21;j++)            dis[i][j]=INF;        scanf("%d",&m);        int start,thend;        printf("Test Set #%d\n",nn++);        while(m--)        {            scanf("%d%d",&start,&thend);            if(!sp[start])            {                SPFA(start);                sp[start]=1;            }            printf("%d to %d: %d\n",start,thend,dis[start][thend]);        }        printf("\n");    }}

Poj 1603 risk