【讲解】1030 Travel Plan (30 分)【DFS】

立志用最少的代码做最高效的表达

PAT甲级最优题解——>传送门

A traveler’s map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination. If such a shortest path is not unique, you are supposed to output the one with the minimum cost, which is guaranteed to be unique.

Input Specification:
Each input file contains one test case. Each case starts with a line containing 4 positive integers N, M, S, and D, where N (≤500) is the number of cities (and hence the cities are numbered from 0 to N−1); M is the number of highways; S and D are the starting and the destination cities, respectively. Then M lines follow, each provides the information of a highway, in the format:

City1 City2 Distance Cost
where the numbers are all integers no more than 500, and are separated by a space.

Output Specification:
For each test case, print in one line the cities along the shortest path from the starting point to the destination, followed by the total distance and the total cost of the path. The numbers must be separated by a space and there must be no extra space at the end of output.

Sample Input:
4 5 0 3
0 1 1 20
1 3 2 30
0 3 4 10
0 2 2 20
2 3 1 20

Sample Output:
0 2 3 3 40

采用vector做dfs的参数，同时求出最短距离和路径。具体逻辑请读者去代码中体会。

#include<bits/stdc++.h>
using namespace std;int G[510][510], vis[510], cost[510][510];  //地图、vis数组、走每条路的花费成本
int n, m, s_c, d_c;                         //城市数、道路数、起点、终点
vector<int>fin;                               //存储最终路线
int min_dis = 0x3f3f3f3f, min_cost = 0x3f3f3f3f;  //最短距离和花费 void dfs(int now_dis, int now_cost, vector<int>v) {int s = v.back();                       //最后一个元素为当前位置 if(now_dis > min_dis) return;          //剪枝if(s == d_c) {                            //边界条件 if(now_dis == min_dis)                 //距离相同则比较花费 if(now_cost > min_cost) return;min_dis = now_dis; min_cost = now_cost;fin = v;return; }for(int i = 0; i < n; i++)               //逐个点遍历 if(vis[i] == 0 && G[s][i]>0) {     //如果没访问过且连通v.push_back(i);              //v数组压入该点 vis[i] = 1;                      //置1表示已访问 dfs(now_dis+G[s][i], now_cost+cost[s][i], v);vis[i] = 0;                       //回溯 v.pop_back();}
}int main() {cin >> n >> m >> s_c >> d_c;for(int i = 0; i < m; i++) {int c1, c2, dis, val; cin >> c1 >> c2 >> dis >> val;G[c1][c2] = G[c2][c1] = dis;       //构建地图和距离 cost[c1][c2] = cost[c2][c1] = val;  //花费 }vector<int>v;  v.push_back(s_c);    //压入起点 vis[s_c] = 1;dfs(0, 0, v);  //距离、花费、路线 for(int i = 0; i < fin.size(); i++) cout << fin[i] << ' ';cout << min_dis << ' ' << min_cost << '\n';return 0;
}

耗时：

【讲解】1030 Travel Plan (30 分)【DFS】_41行代码Ac相关推荐

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【讲解】1030 Travel Plan (30 分)【DFS】_41行代码Ac

立志用最少的代码做最高效的表达

耗时：

【讲解】1030 Travel Plan (30 分)【DFS】_41行代码Ac相关推荐

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