pat-a1072. Gas Station (30)

直接对每个能建加油站的地方使用dijkstra算法就行了，求出到每个用户的最短距离，将每次的结果求出总路程，离最近用户距离，以及加油站下标存放在结构体数组里面然后排序。输出第一组数据就行。

四舍五入那里3.25保留一位我的编译环境会输出3.2，直接提交到oj上也是对的，应该是和编译环境有关。

#include<cstdio>#include<map>#include<vector>#include<cstring>#include<queue>#include<algorithm>using namespace std;int dis[1100];int visit[1100];const int inf=10000000;struct node{int to,p;node(int a=0,int b=0):to(a),p(b){}friend bool operator < (const node& m,const node& n){return m.p>n.p;}};struct ans{int mindis;int sum;int index;ans(int a=0,int b=0,int c=0):mindis(a),sum(b),index(c){}friend bool operator < (const ans& m,const ans& n){if(m.mindis!=n.mindis) return m.mindis>n.mindis;if(m.sum!=n.sum) return m.sum<n.sum;return m.index<n.index;}};ans anss[20];map<int,vector<node> >g;int m,n,ds;void dij(int s){for(int i=1;i<=m+n;++i){dis[i]=inf;visit[i]=0;}dis[s]=0;priority_queue<node> q;q.push({s,dis[s]});while(!q.empty()){node t=q.top();q.pop();int v=t.to;if(visit[v]) continue;visit[v]=1;int len=g[v].size();for(int i=0;i<len;++i){node e=g[v][i];if(dis[e.to]>dis[v]+e.p){dis[e.to]=dis[v]+e.p;q.push({e.to,dis[e.to]});}}}}int fun(int s,char* d){int len=strlen(d);int temp=0;for(int i=s;i<len;++i) temp=temp*10+d[i]-'0';return temp;}int main(){int k;scanf("%d%d%d%d",&n,&m,&k,&ds);char f[3],to[3];int co;for(int i=0;i<k;++i){scanf("%s%s%d",&f,&to,&co);int a=0,b=0;if(f[0]=='G') a=fun(1,f)+n;else a=fun(0,f);if(to[0]=='G') b=fun(1,to)+n;else b=fun(0,to);g[a].push_back({b,co});g[b].push_back({a,co});}int k1=0;for(int i=1;i<=m;++i){dij(i+n);int tm=inf,temp=0,flag=1;for(int i=1;i<=n;++i){if(dis[i]>ds){flag=0;break;}if(tm>dis[i]) tm=dis[i];temp+=dis[i];}if(flag){anss[k1++]={tm,temp,i};}}sort(anss,anss+k1);if(k1){double temp=anss[0].sum*1.0/n;printf("G%d\n%.1f %.1f\n",anss[0].index,anss[0].mindis*1.0,temp);}else printf("No Solution\n");return 0;} 

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 (<= 10³), the total number of houses; M (<= 10), the total number of the candidate locations for the gas stations; K (<= 10⁴), 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 51 2 21 4 21 G1 41 G2 32 3 22 G2 13 4 23 G3 24 G1 3G2 G1 1G3 G2 2

Sample Output 1:

G12.0 3.3

Sample Input 2:

2 1 2 101 G1 92 G1 20

Sample Output 2:

No Solution