PAT (Advanced Level) 1072. Gas Station (30)

枚举一下选的位置，每次算一下就可以了。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<queue>
#include<string>
#include<algorithm>
using namespace std;

const int INF=0x7FFFFFFF;
const int maxn=1000+100;
int n,m,k,ds;
struct Edge
{
    int u,v;
    int dis;
}e[20*maxn];
vector<int>g[maxn];
int tot;
int flag[maxn],dis[maxn];
double ans1,ans2;
int ans;
int MIN;

int get(char *s)
{
    int num=0;
    if(s[0]!='G')
    {
        for(int i=0;s[i];i++)
            num=num*10+s[i]-'0';
    }
    else
    {
        for(int i=1;s[i];i++)
            num=num*10+s[i]-'0';
        num=num+n;
    }
    return num;
}

void SPFA(int s)
{
    queue<int>Q;
    memset(flag,0,sizeof flag);
    for(int i=0;i<=n+m;i++) dis[i]=INF;
    Q.push(s); flag[s]=1; dis[s]=0;
    while(!Q.empty())
    {
        int head=Q.front(); Q.pop(); flag[head]=0;
        for(int i=0;i<g[head].size();i++)
        {
            int id=g[head][i];
            if(dis[head]+e[id].dis<dis[e[id].v])
            {
                dis[e[id].v]=dis[head]+e[id].dis;
                if(flag[e[id].v]==0)
                {
                    flag[e[id].v]=1;
                    Q.push(e[id].v);
                }
            }
        }
    }
}

int main()
{
    scanf("%d%d%d%d",&n,&m,&k,&ds);

    tot=0;
    for(int i=1;i<=k;i++)
    {
        char u[5],v[5]; int dis;
        scanf("%s%s%d",u,v,&dis);
        e[tot].u=get(u),e[tot].v=get(v);
        e[tot].dis=dis;
        g[get(u)].push_back(tot),tot++;

        e[tot].u=get(v),e[tot].v=get(u);
        e[tot].dis=dis;
        g[get(v)].push_back(tot),tot++;
    }

    ans1=1.0*INF;
    ans2=1.0*INF;
    MIN=0;

    for(int i=n+1;i<=n+m;i++)
    {
        SPFA(i);
        int sum=0,MIN_NOW=INF;
        for(int j=1;j<=n;j++)
        {
            if(dis[j]>ds) {sum=-1;break;}
            else
            {
                sum=sum+dis[j];
                MIN_NOW=min(MIN_NOW,dis[j]);
            }
        }
        if(sum==-1) continue;

        if(MIN_NOW>MIN)
        {
            MIN=MIN_NOW;
            ans2=1.0*sum/n;
            ans1=1.0*MIN;
            ans=i;
        }

        else if(MIN_NOW==MIN&&1.0*sum/n<ans2)
        {
            ans2=1.0*sum/n;
            ans=i;
        }
    }
    if(MIN==0) printf("No Solution
");
    else
    {
        printf("G%d
",ans-n);
        printf("%.1lf %.1lf
",ans1,ans2);
    }
    return 0;
}