PAT L3-005. 垃圾箱分布

最短路。

枚举垃圾箱放哪里，然后算最短路。

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

int n,m,k,ds;
char u[];
vector<int>g[];

int v[][];
int dis[], f[];
int fail=;

int ansLen=-,avg=-,id=-;

int get()
{
    int sum=;
    if(u[]!='G')
    {
        for(int i=;u[i];i++) sum=sum*+u[i]-'';
    }
    else
    {
        for(int i=;u[i];i++) sum=sum*+u[i]-'';
        sum = sum + n;
    }
    return sum;
}

void spfa(int x)
{
    for(int i=;i<=n+m;i++) dis[i] = 0x7FFFFFFF;
    queue<int>Q; memset(f,,sizeof f);
    dis[x]=; f[x]=; Q.push(x);

    while(!Q.empty())
    {
        int h= Q.front(); Q.pop(); f[h]=;

        for(int i=;i<g[h].size();i++)
        {
            int to = g[h][i];
            if(dis[h]+v[h][to]<dis[to])
            {
                dis[to] = dis[h]+v[h][to];
                if(f[to]==)
                {
                    f[to]=;
                    Q.push(to);
                }
            }
        }
    }

    for(int i=;i<=n;i++) if(dis[i]>ds) return ;

    fail=;

    int mi=0x7FFFFFFF,sum=;
    for(int i=;i<=n;i++) sum=sum+dis[i],mi = min(mi,dis[i]);

    if(id==-)
    {
        id = x;
        ansLen = mi;
        avg = sum;
    }

    else
    {
        if(mi>ansLen)
        {
            id = x;
            ansLen = mi;
            avg = sum;
        }
        else if(mi==ansLen)
        {
            if(sum<avg)
            {
                id = x;
                ansLen = mi;
                avg = sum;
            }
        }
    }
}

int main()
{
    scanf("%d%d%d%d",&n,&m,&k,&ds);
    for(int i=;i<=k;i++)
    {
        int c;
        scanf("%s",u); int A = get();
        scanf("%s",u); int B = get();
        scanf("%d",&c);

        v[A][B]=v[B][A]=c;
        g[A].push_back(B); g[B].push_back(A);
    }

    for(int i=n+;i<=n+m;i++)
    {
        spfa(i);
    }

    if(fail) printf("No Solution\n");
    else
    {
        printf("G%d\n",id-n);
        printf("%.1f %.1f\n",1.0*ansLen,1.0*avg/n);
    }

    return ;
}