PAT 1072. Gas Station

来源：互联网发布：后醍醐天皇知乎编辑：程序博客网时间：2024/05/24 06:07

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 0x6fffffffint 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

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