HDOJ/HDU 4034 2011成都网络赛D题

最新推荐文章于 2025-12-02 21:20:38 发布

最新推荐文章于 2025-12-02 21:20:38 发布 · 76 阅读

文章标签：

#java

本文探讨了已知各顶点间最短路径长度时如何还原原始图的问题，并通过Floyd算法验证图的合理性，同时介绍了如何计算原图中可能存在的最小边数。

Graph

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 355Accepted Submission(s): 195

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

判断这个图是否合理很简单，就是利用floyd算法在做一次松弛操作。如果发现还有边可以进行松弛的话就输出impossible

然后接下来统计最小边就是去找dis[i][j]能不能表示为dis[i][k]+dis[k][j]的形式

所以说这个题实际上是考察了floyd算法的理解

我的代码：

#include<stdio.h> #include<string.h> int dis[105][105]; int main() { int n,i,j,t,T,k; scanf("%d",&T); for(t=1;t<=T;t++) { scanf("%d",&n); for(i=1;i<=n;i++) for(j=1;j<=n;j++) scanf("%d",&dis[i][j]); bool flag=true; int ans=0; for(k=1;k<=n;k++) for(i=1;i<=n;i++) for(j=1;j<=n;j++) { if(i!=j&&j!=k&&i!=k) { if(dis[i][j]>dis[i][k]+dis[k][j]) { if(flag) printf("Case %d: impossible\n",t); flag=false; } } } if(flag) { for(i=1;i<=n;i++) for(j=1;j<=n;j++) { bool loop=true; if(i!=j) { for(k=1;k<=n;k++) { if(i!=k&&j!=k) { if(dis[i][j]==dis[i][k]+dis[k][j]) loop=false; } } if(loop) ans++; } } printf("Case %d: %d\n",t,ans); } } return 0; }