1072. Gas Station (30) 多源最短路

A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. 

考试的时候这句话意思没有完全理解，悲剧鸟。。。

其实就是让你找一个加油站，它到所有房子最短距离为len , 找一个加油站len的距离越大越好

加油站最多10个，以这几个加油站为原点，做dijk最短路即可

#include<stdio.h>

int map[1099][1099];
const int MAX=499999999;

int use[1099];
int dis[1099];
int n,m,k,ds;
void djik(int start){
    int i,j,end=n+m;
    for(i=0;i<=end;i++){
        use[i]=0;
        dis[i]=MAX;
    }dis[start]=0;

    for(i=1;i<=end;i++){
        int min=MAX,rj;
        for(j=1;j<=end;j++){
            if(use[j]==1)continue;
            if(dis[j]<min){
                min=dis[j];
                rj=j;
            }
        }use[rj]=1;

        for(j=1;j<=end;j++){
            
            if(dis[j]>dis[rj]+map[rj][j])
                dis[j]=dis[rj]+map[rj][j];
        }
    }

    for(i=1;i<=end;i++){
        if(start!=i)
        map[start][i]=dis[i];
    }
}

int main()
{

    while(scanf("%d%d%d%d",&n,&m,&k,&ds)!=EOF){
        int i,j,end,x,ri;
        end=n+m;
        
        for(i=1;i<=end;i++){
            for(j=1;j<=end;j++)map[i][j]=MAX;
        }
        char ss[9],ss2[9];
        int ll,rr,v;
        for(i=1;i<=k;i++){
            scanf("%s",ss);
            scanf("%s",ss2);
            scanf("%d",&v);
            if(ss[0]=='G'){
                if(ss[1]==0){
                    ll=ss[0]-'0'+n;
                }else if(ss[2]==0){
                    ll=ss[1]-'0'+n;
                }else{
                    ll=10+n;
                }
            }else{
                sscanf(ss,"%d",&ll);
            }
            
            if(ss2[0]=='G'){
                if(ss2[1]==0){
                    rr=ss2[0]-'0'+n;
                }else if(ss2[2]==0){
                    rr=ss2[1]-'0'+n;
                }else{
                    rr=10+n;
                }
            }else{
                sscanf(ss2,"%d",&rr);
            }
            
            map[ll][rr]=map[rr][ll]=v;
        }

        for(i=n+1;i<=end;i++){
            djik(i);
        }
        
    /*    for(x=1;x<=end;x++){
            for(i=1;i<=end;i++){
                for(j=1;j<=end;j++){
                    if(map[i][x]+map[x][j]<map[i][j])
                        map[i][j]=map[i][x]+map[x][j];
                }
            }
        }*/
        
        int tempall,all,min=0,fail=0;
        for(i=n+1;i<=end;i++){
            tempall=0;
            int tempmin=MAX;
            int ok=1;
            for(j=1;j<=n;j++){
                if(i==j)continue;
                if(map[i][j]>ds)
                    ok=0;
                if(ok==0)break;
                tempall+=map[i][j];
                
                if(tempmin>map[i][j])
                    tempmin=map[i][j];
            }
            if(ok==0){
                fail++;
                continue;
            }

            if(tempmin==min){
                if(tempall<all){
                    ri=i;
                    all=tempall;
                }
            }
            if(tempmin>min){
                min=tempmin;
                ri=i;
                all=tempall;
            }

        //        printf("G%
        //d
",ri-n);
        //    printf("%.1lf %.1lf
",min*1.0,all*1.0/n);
        }
        
        if(fail==m){
            printf("No Solution
");
        }else{
            printf("G%d
",ri-n);
            printf("%.1lf %.1lf
",min*1.0,all*1.0/n);
        }
    }
    return 0;
}