1111 Online Map（30 分）(cj)

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本文介绍了一个在线地图系统如何推荐从当前位置到目的地的最短和最快路径。通过使用Dijkstra算法来找出满足条件的最佳路径，并详细解释了算法实现过程。

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1111 Online Map（30 分）

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:

Sample Output 1:

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

Sample Input 2:

Sample Output 2:

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

绝望。。。dfs 会超时间 , bfs 会超内存， Ford 多源最短路径不好求路径。。只能用 Dijkstra 单源最短路径了

知识点就是 Dijksta 的路径寻找方式，路径地点个数储存，Dijkstra 的判断条件。

code

#pragma warning(disable:4996)
#include <iostream>
#include <vector>
#define inf 0x7fffffff
using namespace std;
int map[505][505][2];
int distancepath[505][2];
int timepath[505][2];
int path_w[505];
int path_o[505];
int posnum[505];
bool vis[505];
int n, m;
void Dijkstra(int pos) {
	for (int i = 0; i < n; ++i) {
		distancepath[i][0] = map[pos][i][0];
		distancepath[i][1] = map[pos][i][1];
	}
	while (1) {
		int minpos = -1, min_d = 0x7fffffff, min_t = 0x7fffffff;
		for (int i = 1; i < n; ++i) {
			if (vis[i] == 0 && (distancepath[i][0] < min_d || 
				distancepath[i][1] < min_t && 
				distancepath[i][0] == min_d) ) {
				minpos = i;
				min_d = distancepath[i][0];
				min_t = distancepath[i][1];
			}
		}
		vis[minpos] = 1;
		if (minpos == -1) return;
		for (int i = 0; i < n; ++i) {
			if (vis[i] == 0 && (map[minpos][i][0] != inf && distancepath[i][0] > min_d + map[minpos][i][0] ||
				distancepath[i][0] == min_d + map[minpos][i][0] && 
				distancepath[i][1] > min_t + map[minpos][i][1]) ) {
				distancepath[i][0] = min_d + map[minpos][i][0];
				distancepath[i][1] = min_t + map[minpos][i][1];
				path_o[i] = minpos;
			}
		}
	}
}
void Dijkstra_w(int pos) {
	for (int i = 0; i < n; ++i) {
		timepath[i][0] = map[pos][i][1];
		timepath[i][1] = 0;
	}
	while (1) {
		int minpos = -1, min_t = 0x7fffffff;
		for (int i = 1; i < n; ++i) {
			if (vis[i] == 0 && timepath[i][0] < min_t) {
				minpos = i;
				min_t = timepath[i][0];
			}
		}
		vis[minpos] = 1;
		if (minpos == -1) return;
		for (int i = 0; i < n; ++i) {
			if (vis[i] == 0 && (map[minpos][i][1] != inf && timepath[i][0] > min_t + map[minpos][i][1] ||
				timepath[i][0] == min_t + map[minpos][i][1] &&
				timepath[i][1] > timepath[minpos][1] + 1)) {
				timepath[i][0] = min_t + map[minpos][i][1];
				timepath[i][1] = timepath[minpos][1] + 1;
				path_w[i] = minpos;
			}
		}
	}
}
void init() {
	for (int i = 0; i < 505; ++i) {
		path_o[i] = -1;
		path_w[i] = -1;
	}
	for (int i = 0; i < 505; ++i) {
		for (int j = 0; j < 505; ++j) {
			map[i][j][0] = inf;
			map[i][j][1] = inf;
		}
		map[i][i][0] = 0;
		map[i][i][1] = 0;
	}
}
void smallinit() {
	for (int i = 0; i < 505; ++i) {
		vis[i] = 0;
	}
}
int main() {
	init();
	cin >> n >> m;
	int v1, v2, one_way, length, time;
	for (int i = 0; i < m; ++i) {
		cin >> v1 >> v2 >> one_way >> length >> time;
		map[v1][v2][0] = length;
		map[v1][v2][1] = time;
		if (!one_way) {
			map[v2][v1][0] = length;
			map[v2][v1][1] = time;
		}
	}
	int source, destination;
	cin >> source >> destination;
	Dijkstra(source);
	int p = destination;
	int min_d = distancepath[destination][0];
	vector<int> tmp_o,tmp_w;
	while (p != -1) {
		tmp_o.push_back(p);
		p = path_o[p];
	}
	tmp_o.push_back(source);
	smallinit();
	Dijkstra_w(source);
	p = destination;
	int min_t = timepath[destination][0];
	while (p != -1) {
		tmp_w.push_back(p);
		p = path_w[p];
	}
	tmp_w.push_back(source);
	if (tmp_o == tmp_w) {
		cout << "Distance = " << min_d << "; ";
	}
	else {
		cout << "Distance = " << min_d << ": " ;
		for (int i = tmp_o.size()-1; i >= 0; --i) {
			if (i != tmp_o.size()-1) cout << " -> ";
			cout << tmp_o[i];
		}
		cout << endl;
	}
	cout << "Time = " << min_t << ": ";
	for (int i = tmp_w.size()-1; i >= 0; --i) {
		if (i != tmp_w.size()-1) cout << " -> ";
		cout << tmp_w[i];
	}
	cout << endl;
	system("pause");
	return 0;
}