1087. All Roads Lead to Rome (30)

1.求单源最短路径，使用dijkstra求出最小耗费，以这个最小耗费作为约束条件，在后面遍历的时候进行剪枝

AC代码：

//#include<string>//#include <iomanip>//#include<stack>//#include<unordered_set>//#include <sstream>//#include "func.h"//#include <list>#include<unordered_map>#include<set>#include<queue>#include<map>#include<vector>#include <algorithm>#include<stdio.h>#include<iostream>#include<string>#include<memory.h>#include<limits.h>#include<stack>using namespace std;struct cityNode{vector<pair<string, int>> list;bool visited;bool sured;int happiness;int cost;cityNode() :list(0), visited(false), sured(false), happiness(-1), cost(INT_MAX){};};void dfs(string now, int minCost, int nowCost, vector<pair<string, int>>&path, vector<vector<pair<string, int>>>&ans, map<string, bool>&used, map<string, cityNode>&city){if (now == "ROM"&&nowCost == minCost){ans.push_back(path);}else if (nowCost > minCost)return;else{for (int i = 0; i < city[now].list.size(); i++){string q = city[now].list[i].first;if (!used[q]){used[q] = true;path.push_back({ q, city[q].happiness });dfs(q, minCost, nowCost + city[now].list[i].second, path, ans, used, city);path.pop_back();used[q] = false;}}}}bool cmp(const vector<pair<string, int>>&a, const vector<pair<string, int>>&b){int aHappiness = 0;for (int i = 0; i < a.size(); i++){aHappiness += a[i].second;}int bHappiness = 0;for (int i = 0; i < b.size(); i++){bHappiness += b[i].second;}if (aHappiness > bHappiness) return true;else if (aHappiness == bHappiness && aHappiness / a.size() > bHappiness / b.size())return true;else return false;}int main(void){int n, k;string src;cin >> n >> k >> src;string target = "ROM";map<string, cityNode> city;for (int i = 0; i < n - 1; i++){string str;int happiness;cin >> str >> happiness;city[str].happiness = happiness;}for (int i = 0; i < k; i++){string a, b;int cost;cin >> a >> b >> cost;city[a].list.push_back({ b, cost });city[b].list.push_back({ a, cost });}city[src].visited = true;city[src].cost = 0;while (1){string p = "";for (map<string, cityNode>::iterator ite = city.begin(); ite != city.end(); ite++){if (p == ""&&ite->second.visited&&!ite->second.sured)p = ite->first;else if (p != ""&&ite->second.visited&&!ite->second.sured&& ite->second.cost < city[p].cost)p = ite->first;}if (p == "") break;city[p].sured = true;if (city[target].sured) break;for (int i = 0; i < city[p].list.size(); i++){string q = city[p].list[i].first;if (!city[q].sured&&city[p].cost + city[p].list[i].second < city[q].cost){city[q].visited = true;city[q].cost = city[p].cost + city[p].list[i].second;}}}int minCost = city[target].cost;map<string, bool> used;vector<pair<string, int>>path;vector<vector<pair<string, int>>>ans;dfs(src, minCost, 0, path, ans, used, city);sort(ans.begin(), ans.end(), cmp);int totalHappiness = 0;int avgHappiness = 0;for (int i = 0; i < ans[0].size(); i++){totalHappiness += ans[0][i].second;}avgHappiness = totalHappiness / ans[0].size();printf("%d %d %d %d\n", ans.size(), minCost, totalHappiness, avgHappiness);cout << src << "->";for (int i = 0; i < ans[0].size(); i++){cout << ans[0][i].first;if (i != ans[0].size() - 1)cout << "->";}cout << endl;return 0;}