1087 All Roads Lead to Rome (30 分)

题目链接：

题目详情 - 1087 All Roads Lead to Rome (30 分) (pintia.cn)https://pintia.cn/problem-sets/994805342720868352/problems/994805379664297984

题目描述：

Indeed there are many different tourist routes from our city to Rome. You are supposed to find your clients the route with the least cost while gaining the most happiness.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers N (2≤N≤200), the number of cities, and K, the total number of routes between pairs of cities; followed by the name of the starting city. The next N−1 lines each gives the name of a city and an integer that represents the happiness one can gain from that city, except the starting city. Then K lines follow, each describes a route between two cities in the format City1 City2 Cost. Here the name of a city is a string of 3 capital English letters, and the destination is always ROM which represents Rome.

Output Specification:

For each test case, we are supposed to find the route with the least cost. If such a route is not unique, the one with the maximum happiness will be recommanded. If such a route is still not unique, then we output the one with the maximum average happiness -- it is guaranteed by the judge that such a solution exists and is unique.

Hence in the first line of output, you must print 4 numbers: the number of different routes with the least cost, the cost, the happiness, and the average happiness (take the integer part only) of the recommanded route. Then in the next line, you are supposed to print the route in the format City1->City2->...->ROM.

Sample Input:

6 7 HZH
ROM 100
PKN 40
GDN 55
PRS 95
BLN 80
ROM GDN 1
BLN ROM 1
HZH PKN 1
PRS ROM 2
BLN HZH 2
PKN GDN 1
HZH PRS 1

Sample Output:

3 3 195 97
HZH->PRS->ROM

思路：

1. 最短路径的一道题，Dijkstra+DFS， Dijkstra求最短路，DFS打印最短路。

2. 给出起点和终点（终点就是ROM），求最短路，因为城市都是字符串，所以要用map映射为数字方便处理（当然还要做数字到字符串的映射）

3. 求最短路（题中是最小花费）的同时还要求最大的幸福值，所以如果最短路条数不唯一，则还需要看当前路径是否幸福值更大。

4. 因为要打印出最短路径，所以要将S到达ROM的前序路径都保存下来，再通过DFS深度遍历输出路径。

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 202, inf = 0x3f3f3f3f;
// G邻接矩阵,d最小花费,hapi幸福值,num最短路数量,wt最大幸福值,pre保存前序路径
int G[maxn][maxn], d[maxn], hapi[maxn], num[maxn], wt[maxn], pre[maxn];
int n, k, cost, des = 1; // n个城市，k条路，cost花费，des目的地（起点为1）
bool vis[maxn]; // 当前城市是否访问过
vector<string> path; // 保存最终路径
// 将城市名称和数字序列对应
map<string, int> StoI;
map<int, string> ItoS;void Dijkstra(int s){fill(d, d+maxn, inf);    // 初始化d d[s] = 0;  // 到自身的最小花费为0num[s] = 1; // // 到自身的路径只有1条wt[s] = 0; // 到自身的花费为0for(int i = 0; i < n; i++){ // 对n城市都进行一次dijkstraint MIN = inf, u = -1; for(int j = 1; j <= n; j++){ // 查找未访问过的城市中花费最小的if(!vis[j] && d[j] < MIN){MIN = d[j];u = j;}}if(u == -1) return; // 未找到直接返回vis[u] = true; // 标记当前城市已访问for(int j = 1; j <= n; j++){ // 遍历n个城市if(!vis[j] && G[u][j]){ // 能直接到达且未访问过的城市if(d[u] + G[u][j] < d[j]){ // 如果当前路径到达的最小花费更小d[j] = d[u] + G[u][j]; // 更新最小花费路径wt[j] = wt[u] + hapi[j]; // 更新最大幸福值num[j] = num[u]; // 最小花费条数不变pre[j] = u; // 更新前序}else if(d[u] + G[u][j] == d[j]) { // 如果最小花费的路径不唯一num[j] += num[u]; // 最小花费的路径条数累加          if(wt[j] < wt[u] + hapi[j]){ // 如果当前路径最大幸福值更大wt[j] = wt[u] + hapi[j]; //更新最大幸福值pre[j] = u; // 更新前序}} }}}
}void DFS(int d){ // 从终点开始递归if(d == 1){ // 如果到达起点path.push_back(ItoS[d]); // 添加城市return;}DFS(pre[d]); // // 否则递归查找前序路径path.push_back(ItoS[d]); // 添加经过的城市
}int main(){string city; cin >> n >> k >> city; //输入StoI[city] = 1; //起点城市为1ItoS[1] = city;for(int i = 2; i <= n; i++){cin >> city >> hapi[i];StoI[city] = i; //城市名称和数组对应ItoS[i] = city;if(city == "ROM") des = i; // ROM为目的地}for(int i = 0; i < k; i++){string s1, s2; // 城市名称int v1, v2; // 城市序列cin >> s1 >> s2 >> cost;v1 = StoI[s1];v2 = StoI[s2];G[v1][v2] = G[v2][v1] = cost; //赋值v1到v2的花费}Dijkstra(1);DFS(des);printf("%d %d %d %d\n", num[des], d[des], wt[des], wt[des]/(path.size()-1));for(int i = 0; i < path.size(); i++){ //打印路径if(i != 0) printf("->");cout << path[i];}return 0;
}

总结：

这题就是将城市序号变成了字符串，用map映射之后就是一个经典的Dijkstra+DFS的最短路题目，代码基本都是模板，细心点的你一定可以做对。