1111 Online Map (30 分) dijstra最短路+dfs回溯路径

曾经终败给现在 2021-11-01 07:14 299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

**题意：**7个结点，9条边

从x到y，如果k==1则为单向边，随后是长度和花费时间。

输出最短路和最少花费时间

1如果最短路径有多条，输出最快的  
2如果最快路径有多条，输出经过节点数最少的（不是路径最短的！！！！不然测试点2过不去呜呜呜）  
3如果两条路径一样则输出样例2这样的

**思路：**两次dijstra计算最短路和最少花费时间，然后两次dfs回溯路径，先比较再输出。

#include <stdio.h>
    #include <string.h>
    #include <string>
    #include <map>
    #include <vector>
    #include <algorithm>
    #include <iostream>
    #define INF 0x3f3f3f3f
    using namespace std;
    int ma[505][505];
    int cost[505][505];
    int dis[505];
    int sum[505];
    int node[505];
    int vis[505];
    int pre1[505],pre2[505];
    vector<int> path1,path2;
    int n,m,s,e;
    void dfs1(int x)
    {
    	if(x==-1)return;
    	path1.push_back(x);
    	x=pre1[x];
    	dfs1(x);
    }
    void dfs2(int x)
    {
    	if(x==-1)return;
    	path2.push_back(x);
    	x=pre2[x];
    	dfs2(x);
    }
    int main()
    {
    	int i,j,next;
    	scanf("%d%d",&n,&m);
    	memset(ma,INF,sizeof ma);
    	memset(cost,INF,sizeof cost);
    	memset(dis,INF,sizeof dis);
    	memset(sum,INF,sizeof sum);
    	memset(vis,0,sizeof vis);
    	memset(pre1,-1,sizeof pre1);
    	memset(pre2,-1,sizeof pre2);
    	for(i=0;i<m;i++)
    	{
    		int x,y,k,len,ti;
    		scanf("%d%d%d%d%d",&x,&y,&k,&len,&ti);
    		ma[x][y]=len;
    		cost[x][y]=ti;
    		if(k==0)
    		{
    			ma[y][x]=len;
    			cost[y][x]=ti;
    		}
    	}
    	scanf("%d%d",&s,&e);
    	//Distance
    	dis[s]=0;
    	for(i=0;i<n;i++)
    	{
    		int d=INF;
    		for(j=0;j<n;j++)
    		{
    			if(vis[j]==0&&dis[j]<d)
    			{
    				d=dis[j];
    				next=j;
    			}
    		}
    		vis[next]=1;
    		for(j=0;j<n;j++)
    		{
    			if(dis[j]>dis[next]+ma[next][j])
    			{
    				dis[j]=dis[next]+ma[next][j];
    				sum[j]=sum[next]+cost[next][j];
    				pre1[j]=next;
    			}
    			else if(dis[j]==dis[next]+ma[next][j])
    			{
    				if(sum[j]>sum[next]+cost[next][j])
    				{
    					sum[j]=sum[next]+cost[next][j];
    					pre1[j]=next;
    				}
    			}
    		}
    	}
    	//Time
    	memset(sum,INF,sizeof sum);
    	memset(node,INF,sizeof node);
    	memset(vis,0,sizeof vis);
    	sum[s]=0;
    	node[s]=1;
    	for(i=0;i<n;i++)
    	{
    		int d=INF;
    		for(j=0;j<n;j++)
    		{
    			if(vis[j]==0&&sum[j]<d)
    			{
    				d=sum[j];
    				next=j;
    			}
    		}
    		vis[next]=1;
    		for(j=0;j<n;j++)
    		{
    			if(sum[j]>sum[next]+cost[next][j])
    			{
    				sum[j]=sum[next]+cost[next][j];
    				node[j]=node[next]+1;
    				pre2[j]=next;
    			}
    			else if(sum[j]==sum[next]+cost[next][j])
    			{
    				if(node[j]>node[next]+1)
    				{
    					node[j]=node[next]+1;
    					pre2[j]=next;
    				}
    			}
    		}
    	}
    	dfs1(e);
    	dfs2(e);
    	if(path1.size()==path2.size())
    	{
    		int flag=0;
    		for(i=0;i<path1.size();i++)
    		{
    			if(path1[i]!=path2[i])
    			{
    				flag=1;
    				break;
    			}
    		}
    		if(flag)
    		{
    			printf("Distance = %d: %d",dis[e],s);
    			for(j=path1.size()-2;j>=0;j--)
    			printf(" -> %d",path1[j]);
    			printf("\n");
    			printf("Time = %d: %d",sum[e],s);
    			for(j=path2.size()-2;j>=0;j--)
    			printf(" -> %d",path2[j]);
    		}
    		else //路径相同 
    		{
    			printf("Distance = %d; Time = %d: %d",dis[e],sum[e],s);
    			for(j=path1.size()-2;j>=0;j--)
    			printf(" -> %d",path1[j]);
    		}
    	}
    	else
    	{
    		printf("Distance = %d: %d",dis[e],s);
    		for(j=path1.size()-2;j>=0;j--)
    		printf(" -> %d",path1[j]);
    		printf("\n");
    		printf("Time = %d: %d",sum[e],s);
    		for(j=path2.size()-2;j>=0;j--)
    		printf(" -> %d",path2[j]);
    	}
    	printf("\n");
    }