HDU 4034 Graph

系统管理员 2022-08-09 13:46 16阅读 0赞

Problem Description

Everyone knows how to calculate the shortest path in a directed graph. In fact, the opposite problem is also easy. Given the length of shortest path between each pair of vertexes, can you find the original graph?

Input

The first line is the test case number T (T ≤ 100).  
First line of each case is an integer N (1 ≤ N ≤ 100), the number of vertexes.  
Following N lines each contains N integers. All these integers are less than 1000000.  
The jth integer of ith line is the shortest path from vertex i to j.  
The ith element of ith line is always 0. Other elements are all positive.

Output

For each case, you should output “Case k: ” first, where k indicates the case number and counts from one. Then one integer, the minimum possible edge number in original graph. Output “impossible” if such graph doesn't exist.

Sample Input

3  
3  
0 1 1  
1 0 1  
1 1 0  
3  
0 1 3   
4 0 2  
7 3 0  
3  
0 1 4  
1 0 2  
4 2 0

Sample Output

Case 1: 6  
Case 2: 4  
Case 3: impossible

Source

[The 36th ACM/ICPC Asia Regional Chengdu Site —— Online Contest][The 36th ACM_ICPC Asia Regional Chengdu Site _ Online Contest]

Recommend

lcy

题目大意：给定i~j的最短路径，让你求原始图中可能的最少边数。

思路：开始的时候不知道怎么去做尽管老师提示了用floyd去写，可是想到了除了枚举n外并没有想到好办法（但是枚举n不但可能会超时，而且，貌似这种方法是行不通的）！后来看了老师发的题解，又结合网上的资料，才算彻底把问题弄好，正解是 统计一下一条最短路由另外两条最短路构成的数量。然后用所有可能的边减去该数量即可！

code：

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    
    const int N=105;
    int t,tt,n;
    int vis[N][N],mp[N][N];
    
    int solve()
    {
         
    	int i,j,k,m=0;
    	memset(vis,0,sizeof(vis));
    	for(k=0;k<n;k++)
    		for(i=0;i<n;i++)
    			for(j=0;j<n;j++)
    				if(i!=j&&i!=k&&k!=j)
    				{
         
    				    //如果相等则说明ij的连线多余的
    					if(mp[i][j]==mp[i][k]+mp[k][j]&&!vis[i][j])
    					{
         
    						vis[i][j]=1;
    						m++;
    					}
    					//三角形第三边大于两边之和，形不成三角形，构不成图！
    					if(mp[i][j]>mp[i][k]+mp[k][j])
    						return -1;
    				}
    	return m;
    }
    
    int main()
    {
         
    	scanf("%d",&t);
    	for (int ca=1;ca<=t;ca++)
    	{
         
    	    printf("Case %d: ",ca);
    		int i,j;
    		scanf("%d",&n);
    		for(i=0;i<n;i++)
    			for(j=0;j<n;j++)
    				scanf("%d",&mp[i][j]);
            int ans=solve();
    		if  (ans==-1)
    			puts("impossible");
    		else
    			printf("%d\n",n*(n-1)-ans);
    	}
    }

[The 36th ACM_ICPC Asia Regional Chengdu Site _ Online Contest]: http://acm.hdu.edu.cn/search.php?field=problem&key=The%2036th%20ACM/ICPC%20Asia%20Regional%20Chengdu%20Site%20%E2%80%94%E2%80%94%20Online%20Contest&source=1&searchmode=source