团体程序设计天梯赛 L3-007. 天梯地图 Dijkstra

L3-007. 天梯地图

时间限制

300 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

陈越

本题要求你实现一个天梯赛专属在线地图，队员输入自己学校所在地和赛场地点后，该地图应该推荐两条路线：一条是最快到达路线；一条是最短距离的路线。题目保证对任意的查询请求，地图上都至少存在一条可达路线。

输入格式：

输入在第一行给出两个正整数N（2 <= N <=500）和M，分别为地图中所有标记地点的个数和连接地点的道路条数。随后M行，每行按如下格式给出一条道路的信息：

V1 V2 one-way length time

其中V1和V2是道路的两个端点的编号（从0到N-1）；如果该道路是从V1到V2的单行线，则one-way为1，否则为0；length是道路的长度；time是通过该路所需要的时间。最后给出一对起点和终点的编号。

输出格式：

首先按下列格式输出最快到达的时间T和用节点编号表示的路线：

Time = T: 起点 => 节点₁ => ... => 终点

然后在下一行按下列格式输出最短距离D和用节点编号表示的路线：

Distance = D: 起点 => 节点₁ => ... => 终点

如果最快到达路线不唯一，则输出几条最快路线中最短的那条，题目保证这条路线是唯一的。而如果最短距离的路线不唯一，则输出途径节点数最少的那条，题目保证这条路线是唯一的。

如果这两条路线是完全一样的，则按下列格式输出：

Time = T; Distance = D: 起点 => 节点₁ => ... => 终点

输入样例1：

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

输出样例1：

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

输入样例2:

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

输出样例2:

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

#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#include <queue>

using namespace std;

const int maxn = 500, INF = 2147483647;

struct Edge {
	int v1, v2, length, time;
	Edge() {}
	Edge(int v1, int v2, int length, int time) : v1(v1), v2(v2), length(length), time(time) {}
};

struct HeapNode {
	int v, w;
	HeapNode() {}
	HeapNode(int v, int w) : v(v), w(w) {}
	bool operator < (const HeapNode oh) const {
		return oh.w < w;
	}
};

int N, M, s, e;
vector<int> G[maxn + 5], patht, pathd;
vector<Edge> edges;
bool vis[maxn + 5];
int dd[maxn + 5], dt[maxn + 5], dr[maxn + 5], pd[maxn + 5], pt[maxn + 5];
priority_queue<HeapNode> q;

void init1() {
	memset(vis, false, sizeof(vis));
	memset(pt, -1, sizeof(pt));
	while (!q.empty()) q.pop();
	for (int i = 0; i < N; i++) {
		dt[i] = INF;
		dd[i] = INF;
	}
}

void init2() {
	memset(vis, false, sizeof(vis));
	memset(pd, -1, sizeof(pd));
	while (!q.empty()) q.pop();
	for (int i = 0; i < N; i++) {
		dd[i] = INF;
		dr[i] = INF;
	}
}

void dijkstra1() {
	init1();
	q.push(HeapNode(s, 0));
	dt[s] = dd[s] = 0;
	while (!q.empty()) {
		HeapNode thn = q.top(); q.pop();
		int v = thn.v;
		vis[v] = true;
		for (int i = 0; i < G[v].size(); i++) {
			Edge ee = edges[G[v][i]];
			int from = ee.v1, to = ee.v2, length = ee.length, time = ee.time;
			if (!vis[to]) {
				if (dt[from] < INF) {
					if (dt[from] + time < dt[to]) {
						dt[to] = dt[from] + time;
						dd[to] = dd[from] + length;
						pt[to] = from;
						q.push(HeapNode(to, dt[to]));
					}
					else if (dt[from] + time == dt[to] && dd[from] + length < dd[to]) {
						dd[to] = dd[from] + length;
						pt[to] = from;
						q.push(HeapNode(to, dt[to]));
					}
				}
			}
		}
	}
	int x = pt[e];
	patht.push_back(e);
	while (x != -1) {
		patht.push_back(x);
		x = pt[x];
	}
}

void dijkstra2() {
	init2();
	q.push(HeapNode(s, 0));
	dd[s] = dr[s] = 0;
	while (!q.empty()) {
		HeapNode thn = q.top(); q.pop();
		int v = thn.v;
		vis[v] = true;
		for (int i = 0; i < G[v].size(); i++) {
			Edge ee = edges[G[v][i]];
			int from = ee.v1, to = ee.v2, length = ee.length, time = ee.time;
			if (!vis[to]) {
				if (dd[from] < INF) {
					if (dd[from] + length < dd[to]) {
						dd[to] = dd[from] + length;
						dr[to] = dr[from] + 1;
						pd[to] = from;
						q.push(HeapNode(to, dd[to]));
					}
					else if (dd[from] + length == dd[to] && dr[from] + 1 < dr[to]) {
						dr[to] = dr[from] + 1;
						pd[to] = from;
						q.push(HeapNode(to, dd[to]));
					}
				}
			}
		}

	}

	int x = pd[e];
	pathd.push_back(e);
	while (x != -1) {
		pathd.push_back(x);
		x = pd[x];
	}
}

int main()
{
	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		int v1, v2, oneway, length, time;
		scanf("%d%d%d%d%d", &v1, &v2, &oneway, &length, &time);

		edges.push_back(Edge(v1, v2, length, time));
		G[v1].push_back(edges.size() - 1);

		if (!oneway) {
			edges.push_back(Edge(v2, v1, length, time));
			G[v2].push_back(edges.size() - 1);
		}
	}

	scanf("%d%d", &s, &e);
	dijkstra1();
	dijkstra2();
	bool flag = true;
	if (patht.size() == pathd.size()) {
		for (int i = 0; i < patht.size(); i++) {
			if (patht[i] != pathd[i]) {
				flag = false;
				break;
			}
		}
	}
	else {
		flag = false;
	}
	if (flag) {
		printf("Time = %d; ", dt[e]);
		printf("Distance = %d: ", dd[e]);
		for (int i = pathd.size() - 1; i >= 0; i--) {
			printf(i == pathd.size() - 1 ? "%d" : " => %d", pathd[i]);
		}
	}
	else {
		printf("Time = %d: ", dt[e]);
		for (int i = patht.size() - 1; i >= 0; i--) {
			printf(i == patht.size() - 1 ? "%d" : " => %d", patht[i]);
		}
		puts("");
		printf("Distance = %d: ", dd[e]);
		for (int i = pathd.size() - 1; i >= 0; i--) {
			printf(i == pathd.size() - 1 ? "%d" : " => %d", pathd[i]);
		}
	}
	return 0;
}