(PAT 1111) Online Map (Dijkstra+DFS)

一时失言乱红尘 2022-03-19 11:30 135阅读 0赞

Input our current position and a destination, an online map can recommend several paths. Now your job is to recommend two paths to your user: one is the shortest, and the other is the fastest. It is guaranteed that a path exists for any request.

### Input Specification: ###

Each input file contains one test case. For each case, the first line gives two positive integers N (2≤N≤500), and M, being the total number of streets intersections on a map, and the number of streets, respectively. Then M lines follow, each describes a street in the format:

V1 V2 one-way length time

where `V1` and `V2` are the indices (from 0 to N−1) of the two ends of the street; `one-way` is 1 if the street is one-way from `V1` to `V2`, or 0 if not; `length` is the length of the street; and `time` is the time taken to pass the street.

Finally a pair of source and destination is given.

### Output Specification: ###

For each case, first print the shortest path from the source to the destination with distance `D` in the format:

Distance = D: source -> v1 -> ... -> destination

Then in the next line print the fastest path with total time `T`:

Time = T: source -> w1 -> ... -> destination

In case the shortest path is not unique, output the fastest one among the shortest paths, which is guaranteed to be unique. In case the fastest path is not unique, output the one that passes through the fewest intersections, which is guaranteed to be unique.

In case the shortest and the fastest paths are identical, print them in one line in the format:

Distance = D; Time = T: source -> u1 -> ... -> destination

### Sample Input 1: ###

10 15
    0 1 0 1 1
    8 0 0 1 1
    4 8 1 1 1
    3 4 0 3 2
    3 9 1 4 1
    0 6 0 1 1
    7 5 1 2 1
    8 5 1 2 1
    2 3 0 2 2
    2 1 1 1 1
    1 3 0 3 1
    1 4 0 1 1
    9 7 1 3 1
    5 1 0 5 2
    6 5 1 1 2
    3 5

### Sample Output 1: ###

Distance = 6: 3 -> 4 -> 8 -> 5
    Time = 3: 3 -> 1 -> 5

### Sample Input 2: ###

7 9
    0 4 1 1 1
    1 6 1 1 3
    2 6 1 1 1
    2 5 1 2 2
    3 0 0 1 1
    3 1 1 1 3
    3 2 1 1 2
    4 5 0 2 2
    6 5 1 1 2
    3 5

### Sample Output 2: ###

Distance = 3; Time = 4: 3 -> 2 -> 5

解题思路:

题目要求给出两条路径时间最短的一条,路程最短的一条

创建两个二维数组,分别进行两次Dijstra算法,分别是时间最短和距离最短

至于多条最短路径根据条件取其中一条,也要进行两次DFS

本题代码量较大,但题目不难,中规中矩写就能搞定

#define _CRT_SECURE_NO_WARNINGS
    #include <iostream>
    #include <algorithm>
    #include <vector>
    using namespace std;
    const int INF = 0x3fffffff;
    const int MAXV = 510;
    int N, M, DS, DE;
    int STMap[MAXV][MAXV];
    bool STvisited[MAXV];
    int STDis[MAXV], STTime[MAXV];
    int STTimeConsumed[MAXV][MAXV];
    //DFS回溯
    int optsolution = INF;
    int optnodeSolution = INF;
    vector<int> Pres[MAXV];  //先结点
    vector<int> bestPath, tempPath;
    vector<int> bestPathTwo;
    void STDijstra(int s, int WE[],int tmap[MAXV][MAXV]) {
    	fill(WE, WE + MAXV, INF);    //距离为最大
    	fill(STvisited, STvisited + MAXV, false);
    	WE[s] = 0;
    	for (int i = 0; i < N; ++i) {
    		int u = -1, MMIN = INF;
    		for (int j = 0; j < N; ++j) {
    			if (!STvisited[j] && WE[j] < MMIN) {
    				MMIN = WE[j];
    				u = j;
    			}
    		}
    		if (u == -1) return;
    		STvisited[u] = true;
    		for (int v = 0; v < N; ++v) {
    			if (!STvisited[v] && tmap[u][v] != INF) {  //边开放的情况
    				if (tmap[u][v] + WE[u] < WE[v]) {
    					WE[v] = tmap[u][v] + WE[u];
    					Pres[v].clear();
    					Pres[v].push_back(u);
    				}
    				else if (tmap[u][v] + WE[u] == WE[v]) {
    					Pres[v].push_back(u);
    				}
    			}
    		}
    	}
    }
    void STDFS(int vertex,int tmap[MAXV][MAXV]) {
    	if (vertex == DS) {
    		tempPath.push_back(vertex);
    		int temSolution = 0;
    		for (int i = tempPath.size()-1; i > 0;--i) {
    			int curId = tempPath[i], nextId = tempPath[i - 1];
    			temSolution += tmap[curId][nextId];
    		}
    		if (temSolution < optsolution) {
    			optsolution = temSolution;
    			bestPath = tempPath;
    		}
    		tempPath.pop_back();
    		return;
    	}
    	tempPath.push_back(vertex);
    	for (int i = 0; i < Pres[vertex].size(); ++i) {
    		STDFS(Pres[vertex][i],tmap);
    	}
    	tempPath.pop_back();
    }
    
    void STDFS2(int vertex) {
    	if (vertex == DS) {
    		tempPath.push_back(vertex);
    		int nodesize = tempPath.size();
    		if (nodesize < optnodeSolution) {
    			optnodeSolution = nodesize;
    			bestPathTwo = tempPath;
    		}
    		tempPath.pop_back();
    		return;
    	}
    	tempPath.push_back(vertex);
    	for (int i = 0; i < Pres[vertex].size(); ++i) {
    		STDFS2(Pres[vertex][i]);
    	}
    	tempPath.pop_back();
    }
    
    int main() {
    	fill(STMap[0], STMap[0] + MAXV * MAXV, INF); //最大结点
    	fill(STTimeConsumed[0], STTimeConsumed[0] + MAXV * MAXV, INF);
    	scanf("%d %d", &N, &M);
    	int node1, node2, isoneway, nlength, ntime;
    	for (int i = 0; i < M; ++i) {
    		scanf("%d %d %d %d %d", &node1, &node2, &isoneway, &nlength, &ntime);
    		if (isoneway == 1) {
    			STMap[node1][node2] = nlength;
    			STTimeConsumed[node1][node2] = ntime;
    		}
    		else {
    			STMap[node1][node2] = nlength;
    			STTimeConsumed[node1][node2] = ntime;
    			STMap[node2][node1] = nlength;
    			STTimeConsumed[node2][node1] = ntime;
    		}
    	}
    	scanf("%d %d", &DS, &DE);
    	STDijstra(DS, STDis,STMap);
    	STDFS(DE, STTimeConsumed);
    	tempPath.clear();
    	for (int i = 0; i < MAXV; ++i) {
    		Pres[i].clear();
    	}
    	STDijstra(DS, STTime,STTimeConsumed);
    	STDFS2(DE);
    
    	if (bestPath == bestPathTwo) {
    		printf("Distance = %d; Time = %d: ", STDis[DE], STTime[DE]);
    		for (int i = bestPath.size() - 1; i >= 0; --i) {
    			printf("%d", bestPath[i]);
    			if (i > 0) printf(" -> ");
    		}
    	}
    	else {
    		printf("Distance = %d: ", STDis[DE]);
    		for (int i = bestPath.size() - 1; i >= 0; --i) {
    			printf("%d", bestPath[i]);
    			if (i > 0) printf(" -> ");
    		}
    		cout << endl;
    		printf("Time = %d: ", STTime[DE]);
    		for (int i = bestPathTwo.size() - 1; i >= 0; --i) {
    			printf("%d", bestPathTwo[i]);
    			if (i > 0) printf(" -> ");
    		}
    	}
    	system("PAUSE");
    	return 0;
    }