PAT甲级 -- 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:

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

我的大致思路：

应该是要用dij+dfs算法，先求最短的，然后求第二标尺。考虑用结构体，但是中间思路出了点问题，参考了柳神，觉得自己的数据结构还是选择的不合适！

 

 

 柳神代码：

学习点：

1. 还是自己的思路不太清晰，没有完全分析题目，就开始上手敲代码！以后做题要根据题目，列出所求，条件，然后根据特点选择数据结构！

2. 又提供了一种形成路径的新思路

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int inf = 999999999;
int dis[510], Time[510], e[510][510], w[510][510], dispre[510],Timepre[510], weight[510],NodeNum[510];
bool visit[510];
vector<int> dispath, Timepath, temppath;
int st, fin, minnode = inf;
void dfsdispath(int v) {
	dispath.push_back(v);
	if(v == st) return ;
	dfsdispath(dispre[v]);
}
void dfsTimepath(int v) {
	Timepath.push_back(v);
	if(v == st) return ;
	dfsTimepath(Timepre[v]);
}
int main() {
	fill(dis, dis + 510, inf);
	fill(Time, Time + 510, inf);
	fill(weight, weight + 510, inf);
	fill(e[0], e[0] + 510 * 510, inf);
	fill(w[0], w[0] + 510 * 510, inf);
	int n, m;
	scanf("%d %d", &n, &m);
	int a, b, flag, len, t;
	for(int i = 0; i < m; i++) {
		scanf("%d %d %d %d %d", &a, &b, &flag, &len, &t);
		e[a][b] = len;
		w[a][b] = t;
		if(flag != 1) {
			e[b][a] = len;
			w[b][a] = t;
		}
	}
	scanf("%d %d", &st, &fin);
	dis[st] = 0;
	for(int i = 0; i < n; i++)
		dispre[i] = i;
	for(int i = 0; i < n; i++) {
		int u = -1, minn = inf;
		for(int j = 0; j < n; j++) {
			if(visit[j] == false && dis[j] < minn) {
				u = j;
				minn = dis[j];
			}
		}
		if(u == -1) break;
		visit[u] = true;
		for(int v = 0; v < n; v++) {
			if(visit[v] == false && e[u][v] != inf) {
				if(e[u][v] + dis[u] < dis[v]) {
					dis[v] = e[u][v] + dis[u];
					dispre[v] = u;
					weight[v] = weight[u] + w[u][v];
				} else if(e[u][v] + dis[u] == dis[v] && weight[v] > weight[u] + w[u][v]) {
					weight[v] = weight[u] + w[u][v];
					dispre[v] = u;
				}
			}
		}
	}
	dfsdispath(fin);
	Time[st] = 0;
	fill(visit, visit + 510, false);
	for(int i = 0; i < n; i++) {
		int u = -1, minn = inf;
		for(int j = 0; j < n; j++) {
			if(visit[j] == false && minn > Time[j]) {
				u = j;
				minn = Time[j];
			}
		}
		if(u == -1) break;
		visit[u] = true;
		for(int v = 0; v < n; v++) {
			if(visit[v] == false && w[u][v] != inf) {
				if(w[u][v] + Time[u] < Time[v]) {
					Time[v] = w[u][v] + Time[u];
					Timepre[v]=(u);
					NodeNum[v]=NodeNum[u]+1;
				} else if(w[u][v] + Time[u] == Time[v]&&NodeNum[u]+1<NodeNum[v]) {
					Timepre[v]=(u);
					NodeNum[v]=NodeNum[u]+1;
				}
			}
		}
	}
	dfsTimepath(fin);
	printf("Distance = %d", dis[fin]);
	if(dispath == Timepath) {
		printf("; Time = %d: ", Time[fin]);
	} else {
		printf(": ");
		for(int i = dispath.size() - 1; i >= 0; i--) {
			printf("%d", dispath[i]);
			if(i != 0) printf(" -> ");
		}
		printf("\nTime = %d: ", Time[fin]);
	}
	for(int i = Timepath.size() - 1; i >= 0; i--) {
		printf("%d", Timepath[i]);
		if(i != 0) printf(" -> ");
	}
	return 0;
}