PAT 1072. Gas Station

题目：http://pat.zju.edu.cn/contests/pat-a-practise/1072

题解：

对每个加油站进行迪杰斯特拉处理，求出最小距离和平均距离，再按题目要求排序输出。

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<algorithm>
using namespace std;
#define INF 0x6fffffff
int n,m,k,ds;
int mapx[1015][1015];
int distances[1015];
bool flag[1015];
struct point
{
    int id;
    double aver;
    double minx;
    point(int id,double minx,double aver)
    {
        this->id=id;
        this->aver=aver;
        this->minx=minx;
    }
};
vector<struct point> out;
bool cmp(const struct point &a,const struct point &b)
{
    return a.minx==b.minx?(a.aver==b.aver?(a.id<b.id):(a.aver<b.aver)):(a.minx>b.minx);
}
void dijkstra(int x)
{
    int idx,minx;
    for(int i=0; i<n+m; ++i)
    {
        distances[i]=mapx[x][i];
        flag[i]=false;
    }
    for(int i=0; i<n+m; ++i)
    {
        minx=INF;
        for(int j=0; j<n+m; ++j)
        {
            if(distances[j]<minx&&flag[j]==false)
            {
                minx=distances[j];
                idx=j;
            }
        }
        flag[idx]=true;
        for(int j=0; j<n+m; ++j)
            if(mapx[idx][j]+distances[idx]<distances[j])
                distances[j]=mapx[idx][j]+distances[idx];
    }
    double minDist=19999999,total=0;
    for(int i=0; i<n; ++i)
    {
        if(i!=x)
        {
            if(distances[i]>ds)
                return;
            if(distances[i]<minDist)
                minDist=distances[i];
            total+=distances[i];
        }
    }
    struct point z(x,minDist,total/n);
    out.push_back(z);
}
int check(char ch[])
{
    int id=0,x;
    int len=strlen(ch);
    if(ch[0]=='G')
        x=1;
    else
        x=0;
    for(; x<len; ++x)
        id=id*10+ch[x]-'0';
    if(ch[0]=='G')
        id+=n;
    return id-1;
}
int main()
{
    char a[5],b[5];
    int dist;
    int x,y;
    scanf("%d%d%d%d",&n,&m,&k,&ds);
    for(int i=0; i<n+m; ++i)
        for(int j=0; j<n+m; ++j)
        {
            if(i==j)
                mapx[i][j]=0;
            else
                mapx[i][j]=INF;
        }
    for(int i=0; i<k; ++i)
    {
        scanf("%s%s%d",a,b,&dist);
        x=check(a);
        y=check(b);
        mapx[x][y]=mapx[y][x]=dist;
    }
    for(int i=n; i<n+m; ++i)
        dijkstra(i);
    if(!out.empty())
    {
        sort(out.begin(),out.end(),cmp);
        printf("G%d\n%.1f %.1f\n",out[0].id-n+1,out[0].minx,out[0].aver);
    }
    else
    {
        printf("No Solution\n");
    }
    return 0;
}

来源： http://blog.csdn.net/acm_ted/article/details/20881645