1072. Gas Station 解析

这个题的题目真的是好长好长，而且要理清的东西挺多的。不过后面的输出给了比较时候要用的数据的提示。

首先，我们要保证加油站到居民楼越远越好，靠的是min{加油站到剧名楼的距离}的大小，取大的。

然后min{}还相等，取avg{加油站到剧名楼的距离}的大小，取小的。

然后avg{}还相等，取编号小的。

这个题为了便于处理，我把编号连续化了，居民楼的编号排前面，加油站的接在后面。这样dijstra的算法能好写点。

理清了上面，后面都是传统的Dijstra算法了。。。

#include <iostream>#include <algorithm>#include <cstring>#include <climits>#include <string>#include <map>#include <set>#include <queue>#include <stack>#include "float.h"#define MAX 11111using namespace std;int n,m,k,ds;struct node{int ed;float dis;};vector <node> g[MAX];//float dis[MAX];bool isVis[MAX];struct ansNode{float avg;int no;float min;ansNode(){avg = FLT_MAX;};};ansNode ans;int str2int(string s){int pos = 0;int ans = 0;int gasNum = 0;if(s[0] == 'G'){pos = 1;gasNum = n;}for(;pos < s.size() ;pos++){ans *= 10;ans += int(s[pos] - '0');}return ans + gasNum;}void Init(){for(int i = 0 ;i < MAX ;i++){dis[i] = FLT_MAX;isVis[i] = false;}}void Dijstra(int root){dis[root] = 0;for(int i = 1 ; i <= n + m ;i++){int min = INT_MAX;int u = -1;for(int j = 1 ; j <= n+m ;j++){if(!isVis[j] && dis[j] < min){min = dis[j];u = j;}}if(u == -1)return;isVis[u] = true;for(int j = 0;j < g[u].size() ;j++){int v = g[u][j].ed;if(dis[v] > dis[u] + g[u][j].dis)dis[v] = dis[u] + g[u][j].dis;}}}void printfNo(int a){cout << "G" << a-n << endl;}int main(){scanf("%d%d%d%d",&n,&m,&k,&ds);string st,ed;int d;node t;for(int i = 0 ;i < k ;i++){cin >> st >> ed >> t.dis;int n_ed = str2int(ed);int n_st = str2int(st);t.ed = n_ed;g[n_st].push_back(t);t.ed = n_st;g[n_ed].push_back(t);}bool haveAns = false;for(int i = n + 1 ;i <= n+m ; i++){Init();Dijstra(i);float min = FLT_MAX;float Sum = 0;float avg;bool tag = true;for(int j = 1 ; j <= n ;j++){if(dis[j] > ds){tag = false;break;}Sum += dis[j];if(min > dis[j])min = dis[j];}if(tag){avg = Sum/float(n);haveAns = true;if(ans.min < min){ans.no = i;ans.avg = avg;ans.min = min;}else if(ans.min == min && ans.avg > avg){ans.no = i;ans.avg = avg;ans.min = min;}else if(ans.min == min && ans.avg == avg && ans.no > i){ans.no = i;ans.avg = avg;ans.min = min;}}}if(haveAns){printfNo(ans.no);printf("%.01f %.01f\n",ans.min,ans.avg);}elseprintf("No Solution\n");return 0;}