HDOJ 题目4034 Graph（逆向Floyd）

最新推荐文章于 2018-11-27 20:45:48 发布

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最新推荐文章于 2018-11-27 20:45:48 发布

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分类专栏：最短路径

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本文探讨了在给定每个顶点对之间的最短路径长度的情况下，如何重建原始有向图的问题。通过输入数据，我们展示了如何确定可能的边数，并在特定案例中给出了解决方案。

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Graph

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 1992 Accepted Submission(s): 996

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

Sample Output

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

Source

The 36th ACM/ICPC Asia Regional Chengdu Site —— Online Contest

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floyd的松弛部分是 g[i][j] = min(g[i][j], g[i][k] + g[k][j]);也就是说，g[i][j] <= g[i][k] + g[k][j] (存在i->j, i->k, k->j的边)。

ac代码

#include<stdio.h>
#include<string.h>
int map[1010][1010],v[1010][1010];
int main()
{
	int t,c=0,k;
	scanf("%d",&t);
	while(t--)
	{
		int n,i,j,ans,w=1;
		scanf("%d",&n);
		for(i=0;i<n;i++)
		{
			for(j=0;j<n;j++)
				scanf("%d",&map[i][j]);
		}
		ans=n*(n-1);
		memset(v,0,sizeof(v));
		for(k=0;k<n;k++)
		{
			for(i=0;i<n;i++)
			{
				for(j=0;j<n;j++)
				{
					if(k==i||j==k)
						continue;
					if(map[i][j]>map[i][k]+map[k][j])
					{
						w=0;
						break;
					}
					if(!v[i][j]&&map[i][j]==map[i][k]+map[k][j])
					{
						ans--;
						v[i][j]=1;
					}
				}
				if(!w)
					break;
			}
			if(!w)
				break;
		}
		if(w)
			printf("Case %d: %d\n",++c,ans);
		else
			printf("Case %d: impossible\n",++c);
	}
}