【笨方法学PAT】1030 Travel Plan （30 分）

最新推荐文章于 2021-02-26 09:28:25 发布

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最新推荐文章于 2021-02-26 09:28:25 发布

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本文介绍了一种算法，用于解决旅行者寻找从起点到目的地的最短路径及最低成本的问题。通过深度优先搜索（DFS），算法遍历所有可能路径，找出距离最短且成本最低的路线。

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一、题目

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:

Sample Output:

0 2 3 3 40

二、题目大意

求最小路径和最小代价，并输出路径。

三、考点

DFS

四、注意

1、使用fill对数组和二维数组进行初始化。

五、代码

#include<iostream>
#include<vector>
#define N 505
#define INF 9999999
using namespace std;
int e[N][N],c[N][N];
bool visit[N] = { false };
vector<int> path,out_path;
int n, m, s, d, min_dis=INF, min_cost=INF;

void dfs(int index,int dis,int cost) {
	if (index == d) {
		if (min_dis > dis || min_dis == dis&&cost < min_cost) {
			min_dis = dis;
			out_path.clear();
			out_path = path;
			min_cost = cost;
		}
		return;
	}

	for (int i = 0; i < n; ++i) {
		if (e[index][i] != INF&&visit[i] == false) {
			visit[i] = true;
			path.push_back(i);
			dfs(i, dis + e[index][i], cost + c[index][i]);
			visit[i] = false;
			path.pop_back();
		}
	}
	return;
}

int main() {
	//read
	cin >> n >> m >> s >> d;

	fill(e[0], e[0] + N*N, INF);
	fill(visit, visit + n, false);

	for (int i = 0; i < m; ++i) {
		int c1, c2, dis, cost;
		cin >> c1 >> c2 >> dis >> cost;
		e[c1][c2] = e[c2][c1] = dis;
		c[c1][c2] = c[c2][c1] = cost;
	}

	//dfs
	visit[s] = true;
	path.push_back(s);
	dfs(s, 0, 0);

	//output
	for (int i = 0; i < out_path.size(); ++i) {
		if (i != 0)
			cout << " ";
		cout << out_path[i];
	}
	cout << " " << min_dis << " " << min_cost << endl;

	system("pause");
	return 0;
}