1072 Gas Station (30)

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. However it must guarantee that all the houses are in its service range.

Now given the map of the city and several candidate locations for the gas station, you are supposed to give the best recommendation. If there are more than one solution, output the one with the smallest average distance to all the houses. If such a solution is still not unique, output the one with the smallest index number.

Input Specification:

Each input file contains one test case. For each case, the first line contains 4 positive integers: N (<= 10^3^), the total number of houses; M (<= 10), the total number of the candidate locations for the gas stations; K (<= 10^4^), the number of roads connecting the houses and the gas stations; and D~S~, the maximum service range of the gas station. It is hence assumed that all the houses are numbered from 1 to N, and all the candidate locations are numbered from G1 to GM.

Then K lines follow, each describes a road in the format\ P1 P2 Dist\ where P1 and P2 are the two ends of a road which can be either house numbers or gas station numbers, and Dist is the integer length of the road.

Output Specification:

For each test case, print in the first line the index number of the best location. In the next line, print the minimum and the average distances between the solution and all the houses. The numbers in a line must be separated by a space and be accurate up to 1 decimal place. If the solution does not exist, simply output “No Solution”.

Sample Input 1:

4 3 11 5
1 2 2
1 4 2
1 G1 4
1 G2 3
2 3 2
2 G2 1
3 4 2
3 G3 2
4 G1 3
G2 G1 1
G3 G2 2

Sample Output 1:

G1
2.0 3.3

Sample Input 2:

2 1 2 10
1 G1 9
2 G1 20

Sample Output 2:

No Solution

//DEV-C++ 上第一组数据ansAvg == 3.2 C-free ansAvg == 3.3 应该是double的取近似数的判断不同 
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 10020; // maxn = 100010  后面有个的d[j] < minDis 会出现数组溢出。 
const int INF = 100000000;

int n,m,k,DS;
int G[maxn][maxn],d[maxn];
bool vis[maxn] = {false};

void Dijkstra(int s){

    fill(d,d+maxn,INF);
    memset(vis,false,sizeof(vis));
    d[s] = 0;
    for(int i = 0; i < n+m; i++){
        int u = -1,MIN = INF;
        for(int j = 1; j <= n+m; j++){   // 错误1 
            if(vis[j] == false && d[j] < MIN){
                u = j;
                MIN = d[j];
            }
        }
        if( u == -1) return;
        vis[u] = true;
        for(int v = 1; v <= n+m; v++){
            if(vis[v] == false && G[u][v] != INF){
                if(d[v] > d[u] + G[u][v]){
                    d[v] = d[u] + G[u][v];
                }
            }
        }
    }
}

int getID(char str[]){
    int i=0,len = strlen(str),ID = 0; //错误2 
    while(i < len){
        if(str[i] != 'G'){
            ID = ID * 10 + (str[i] - '0');
        }
        i++;
    }
    if(str[0] == 'G') return n+ID;
    else return ID;
}

int main(){
    scanf("%d%d%d%d",&n,&m,&k,&DS);
    fill(G[0],G[0]+maxn*maxn,INF);
    int u,v,w;
    char city1[5],city2[5];
    for(int i = 0; i < k; i++){
        scanf("%s %s %d",city1,city2,&w);
        u = getID(city1);
        v = getID(city2);
        G[u][v] = G[v][u] = w;
    }
    // ansDis是找最短距离中最大的那个，初始值要最小 
    double ansDis = -1, ansAvg = INF;  
    int ansID = -1;
    for(int i = n+1; i <= n+m; i++){
        //minDis是找所有点中的最短距离，初始值最大 
        double minDis = INF, avg = 0;
        Dijkstra(i);
        for(int j = 1; j <= n; j++){  //遍历所有居民点，1-n 
            if(d[j] > DS){
                minDis = -1;
                break;
            }
            if(d[j] < minDis) minDis = d[j];
            avg += 1.0 * d[j] / n;
        }
        if(minDis == -1) continue;
        if(minDis > ansDis){
            ansID = i;
            ansDis = minDis;
            ansAvg = avg;
        }else if(minDis == ansDis && ansAvg > avg){
            ansID = i;
            ansAvg =avg;
        }
    }
    if(ansID == -1) printf("No Solution");
    else{
        printf("G%d\n",ansID - n);
        printf("%.1f %.1f",ansDis,ansAvg);
    }
    return 0;
}