【PAT】1030. Travel Plan (30)

最新推荐文章于 2021-03-12 19:34:52 发布

realxuejin

最新推荐文章于 2021-03-12 19:34:52 发布

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分类专栏： PAT 文章标签： PAT 最短路

本文链接：https://blog.youkuaiyun.com/realxuejin/article/details/49474343

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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

分析：多条最短路，找出最优的最短路。和1018那道题目相似：http://blog.youkuaiyun.com/realxuejin/article/details/49453677

#include <iostream>
#include <vector>
using namespace std;
#define INF 999999
vector<vector<int> >cost;
vector<vector<int> >dist;
vector<bool> used;
vector<vector<int> > allPath;
int cityNum, wayNum, s, des;
vector<int> d;
vector<int> possiblePath;
vector<int> bestPath;
int minCost = INF;
void findBestPath(int u){ 
	possiblePath.push_back(u);
	if(u==s){
		int costTmp=0;
		int i;
		for(i=possiblePath.size()-1; i>0; i--){
			int from = possiblePath[i];
			int to = possiblePath[i-1];
			costTmp += cost[from][to];
		}		
		if(costTmp<minCost){
			minCost = costTmp;
			bestPath = possiblePath;
		}
		return;		
	}
	
	for(int i=0; i<allPath[u].size(); i++){
		findBestPath(allPath[u][i]);
		possiblePath.pop_back();
	}	
}

void dijkstra(int s){
	cost[s][s] = 0;
	allPath.resize(cityNum);	
	while(true){
		int v = -1;
		//从尚未使用过的顶点中选择一个距离最小的顶点
		for(int u=0; u<cityNum; u++){
			if(!used[u] && (v==-1 || dist[s][u]<dist[s][v] ))
				v = u;
		} 
		if(v == -1)
		 	break;
		used[v] = true;
		for(int u=0; u<cityNum; u++){
			if(d[u] > d[v] + dist[v][u] ){
				allPath[u].clear();
				allPath[u].push_back(v);
				d[u] = d[v] + dist[v][u];
			}else if(d[u] == d[v] + dist[v][u]){
				allPath[u].push_back(v);
			}
		}
	}
	return;
}

int main(int argc, char** argv) {
	scanf("%d%d%d%d",&cityNum, &wayNum, &s, &des);
	used.resize(cityNum,false);
	int i;
	int city1, city2, distTmp, costTmp;
    cost.resize(cityNum, vector<int>(cityNum,INF) );
	dist.resize(cityNum, vector<int>(cityNum,INF));
	
	for(i=0; i<wayNum; i++){
		scanf("%d%d%d%d",&city1,&city2,&distTmp,&costTmp);
		dist[city1][city2] = dist[city2][city1] = distTmp;
		cost[city1][city2] = cost[city2][city1] = costTmp;		
	}
	d.resize(cityNum,INF);
	d[s] = 0;
	for(i=0; i<cityNum; i++){
		if(dist[s][i] != INF){
			d[i] = dist[s][i];
		}
	}
	
	dijkstra(s);
	findBestPath(des);
	
	for(i=bestPath.size()-1; i>=0; i--){
		printf("%d ",bestPath[i]);
	}
	printf("%d %d\n",d[des],minCost);
	
	return 0;
}