天梯地图 （30 分）

7-15 天梯地图 （30 分）

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

 代码如下。2份代码 仅打印路径不同。

#include<iostream>
#include<cstring>
#include<math.h>
using namespace std;
const int INF = 0x3f3f3f3f;
int N, E;
int len[505][505], Time[505][505];;
int time_d[505];
int dist[505];
bool vis[505];
int tpath[505], dpath[505];
int num[505];
int a, b;

void tDijskstra(int x){
	memset(time_d, INF, sizeof(time_d));
	memset(dist, INF, sizeof(dist));
	memset(tpath, -1, sizeof(tpath));
	memset(vis, 0, sizeof(vis));
	time_d[x] = 0;
	dist[x] = 0;
	while (1){
		int k = -1, minn = INF;
		for(int i = 0; i < N; i++){
			if( !vis[i] && minn > time_d[i]){
				minn = time_d[i];
				k = i;
			}
		}
		if(k == -1) break;
		vis[k] = 1;
		for(int i = 0; i < N;i++){
			if(!vis[i]){
				if(time_d[i] > time_d[k] + Time[k][i]){
					time_d[i] = time_d[k] + Time[k][i];
					dist[i] = dist[k] + len[k][i]; 
					tpath[i] = k;
				}
				else if(time_d[i] == time_d[k] + Time[k][i]){
					if(dist[i] > dist[k] + len[k][i]){
						dist[i] = dist[k] + len[k][i];
						tpath[i] = k;
					}
				}
			}
		}
	} 
}
void dDijskstra(int x){
	memset(num, 0, sizeof(num));
	memset(dist, INF, sizeof(dist));
	memset(dpath, -1, sizeof(dpath));
	memset(vis, 0, sizeof(vis));
	
	dist[x] = 0;
	while (1){
		int k = -1, minn = INF;
		for(int i = 0; i < N; i++){
			if( !vis[i] && minn > dist[i]){
				minn = dist[i];
				k = i;
			}
		}
		if(k == -1) break;
		vis[k] = 1;
		for(int i = 0; i < N;i ++){
			if(!vis[i]){
				if(dist[i] > dist[k] + len[k][i]){
					dist[i] = dist[k] + len[k][i];
					num[i] = num[k] + 1; 
					dpath[i] = k;
				}
				else if(dist[i] == dist[k] + len[k][i]){
					if(num[i] > num[k] + 1){
						num[i] = num[k] + 1;
						dpath[i] = k;
					}
				}
			}
		}
	} 
}
int judge(int b){
	if(tpath[b] != dpath[b]){
		return 0;
	}
	else if(b != -1){
		return judge(tpath[b]);
	}
	else{		
		return 1;
	}
}
void print(int path[], int x){
	if(x == -1){
		return ;
		
	}
	else{
		print(path, path[x]); 
		cout << x << " => ";
	}
}
void show(int flag){
	if(flag){
		cout << "Time = " << time_d[b] << "; ";
		cout << "Distance = " << dist[b] << ": ";
		print(dpath, dpath[b]);
		cout << b;
	}
	else{
		cout << "Time = " << time_d[b] << ": ";
		print(tpath, tpath[b]);
		cout << b << endl;
		cout << "Distance = " << dist[b] << ": ";
		print(dpath, dpath[b]);
		cout << b; 
	}
}
int main(){
	cin >> N >> E;
	memset(Time, INF, sizeof(Time));
	memset(len, INF, sizeof(len)); 
	
	for(int i = 1; i <= E; i++){
		int v1, v2, one, l, t;
		cin >> v1 >> v2 >> one >> l >> t;
		if(one == 1){
			Time[v1][v2] = min(Time[v1][v2], t);
			len[v1][v2] = min(len[v1][v2], l);
		}
		else{
			Time[v2][v1] = Time[v1][v2] = min(Time[v1][v2], t);
			len[v2][v1] = len[v1][v2] = min(len[v1][v2], l);
		}
	}
	cin >> a >> b;
	tDijskstra(a);
	dDijskstra(a);
	
	int flag = judge(b);
	show(flag);
	return 0;
}

 

#include<iostream>
#include<cstring>
#include<math.h>
#include<stdio.h>
#include<algorithm> 
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
const int INF = 0x3f3f3f3f;
int N, E;
int a, b;
int mp[505][505];
int tim[505][505];
bool vis[505];
int time_d[505];
int dist[505];
int distpath[505];
int timepath[505];
int num[505];


void show(int y){
	stack<int> sta1;
	stack<int> sta2;
	string s1 = "";
	string s2 = "";
	int temp1 = timepath[y];
	int temp2 = distpath[y];
	
	sta1.push(y);
	s1 += y + '0';
	sta2.push(y);
	s2 += y + '0';
	
	while(temp1 != -1){
		s1 += temp1 + '0';
		sta1.push(temp1);
		temp1 = timepath[temp1];
	}
	while(temp2 != -1){
		s2 += temp2 + '0';
		sta2.push(temp2);
		temp2 = distpath[temp2];
	}
	if(s1 == s2){
		cout << "Time = " << time_d[y] << "; "; 
		cout << "Distance = " << dist[y] << ": ";
		cout << sta2.top() ;
		sta2.pop();
		while(!sta2.empty()){
			cout << " => " << sta2.top();
			sta2.pop();
		} 
	}
	else
	{
		cout << "Time = " << time_d[y] << ": "; 
		cout << sta1.top();
		sta1.pop();
		while(!sta1.empty()){
			cout << " => " << sta1.top();
			sta1.pop();
		}
		cout << endl;
		cout << "Distance = " << dist[y] << ": "; 
		cout << sta2.top();
		sta2.pop();
		while(!sta2.empty()){
			cout << " => " << sta2.top();
			sta2.pop();
		}
	}
}
void timeDijskstra(int x){
	memset(vis, 0, sizeof(vis));
	memset(dist, INF, sizeof(dist));
	memset(time_d, INF, sizeof(time_d));
	
	time_d[x] = 0;
	timepath[x] = -1;
	dist[x] = 0;
	while(1){
		int k = -1, minn = INF;
		for(int i = 0; i < N; i++){
			if(!vis[i] && minn > time_d[i]){
				k = i;
				minn = time_d[i];
			}
		}
		if(k == -1) break;
		
		vis[k] = 1;
		for(int i = 0; i < N; i++){
			if(!vis[i]){
				if(time_d[i] > tim[k][i] + time_d[k]){
					dist[i] = mp[k][i] + dist[k];
					timepath[i] = k;
					time_d[i] = time_d[k] + tim[k][i];
				}
				else if(time_d[i] == tim[k][i] + time_d[k]){
					if(dist[i] > mp[k][i] + dist[k]){
						timepath[i] = k;
						dist[i] = mp[k][i] + dist[k];
					}
				}
			}
		}
	}
}
void disDijskstra(int x){
	memset(vis, 0, sizeof(vis));
	memset(dist, INF, sizeof(dist));
	memset(num, INF, sizeof(num));
	
	dist[x] = 0;
	while(1){
		int k = -1, minn = INF;
		for(int i = 0; i < N; i++){
			if(!vis[i] && minn > dist[i]){
				k = i;
				minn = dist[i];
			}
		}
		if(k == -1) break;
		vis[k] = 1;
		for(int i = 0; i < N; i++){
			if(!vis[i]){
				if(dist[i] > dist[k] + mp[k][i]){
					dist[i] = dist[k] + mp[k][i];
					distpath[i] = k;
					num[i] = num[k] + 1;
				}
				else if(dist[i] == dist[k] + mp[k][i]){
					if(num[i] > num[k] + 1){
						num[i] = num[k] + 1;
						distpath[i] = k;
					}
				}
			}
		}
	}	
	show(b);
}

int main(){
	cin >> N >> E;
	memset(mp, INF, sizeof(mp));
	memset(tim, INF, sizeof(tim));
	memset(distpath, -1, sizeof(distpath));
	memset(timepath, -1, sizeof(timepath));
	
	for(int i = 0; i < E; i++){
		int way, l, t, v1, v2; 
		cin >> v1 >> v2 >> way >> l >> t; 
		if(way == 1){
			mp[v1][v2] = min(mp[v1][v2], l);
			tim[v1][v2] = min(tim[v1][v2], t);
		}
		else {
			mp[v1][v2] = mp[v2][v1] = min(mp[v1][v2], min(mp[v2][v1], l));
			tim[v1][v2] = tim[v2][v1] = min(tim[v1][v2], min(tim[v2][v1], t));
		}
	}
	cin >> a >> b;
	
	timeDijskstra(a);
	disDijskstra(a);

	return 0;
}