There is an easy way to obtain a new task from an old one called "Inverse the problem": we give an output of the original task, and ask to generate an input, such that solution to the original problem will produce the output we provided. The hard task of Topcoder Open 2014 Round 2C, InverseRMQ, is a good example.
Now let's create a task this way. We will use the task: you are given a tree, please calculate the distance between any pair of its nodes. Yes, it is very easy, but the inverse version is a bit harder: you are given an n × n distance matrix. Determine if it is the distance matrix of a weighted tree (all weights must be positive integers).
The first line contains an integer n (1 ≤ n ≤ 2000) — the number of nodes in that graph.
Then next n lines each contains n integers di, j (0 ≤ di, j ≤ 109) — the distance between node i and node j.
If there exists such a tree, output "YES", otherwise output "NO".
3 0 2 7 2 0 9 7 9 0
YES
3 1 2 7 2 0 9 7 9 0
NO
3 0 2 2 7 0 9 7 9 0
NO
3 0 1 1 1 0 1 1 1 0
NO
2 0 0 0 0
NO
In the first example, the required tree exists. It has one edge between nodes 1 and 2 with weight 2, another edge between nodes 1 and 3 with weight 7.
In the second example, it is impossible because d1, 1 should be 0, but it is 1.
In the third example, it is impossible because d1, 2 should equal d2, 1.
给定一个矩阵，表示每两个节点之间的权值距离，问能否够相应生成一棵树，
使得这棵树中的随意两点之间的距离和矩阵中的相应两点的距离相等。
思路：我们将给定的矩阵看成是一个图，a 到 b会有多条路径， 假设存在一棵树。那么
这个树中a->b的距离一定是这个图中全部a->b中路径长度最短的一条！
所以我们依据边权，
建立一棵MST树！
再将MST树中的随意两点之间的距离求出来，看是否和矩阵中的相应的节点
对距离同样,我们首先构造一个最小生成树,然后比較各各点之间的距离是否与题目给出的距离相等,能够用dfs搜索整张图的每两个点之间的距离.以下给的做法非dfs做的,用一个数组f[][],记录x,y两点之间的距离,算距离的时候是通过眼下点的前驱找,也就是说须要一个数组记录前驱,这样就能够不用dfs了,直接能够算.看完代码就明确了.....
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 2010;
const int INF = 0x3f3f3f3f;
int graph[maxn][maxn];
int prior[maxn];
int visit[maxn];
int dis[maxn];
int f[maxn][maxn];
int n;
bool check()
{
for(int i = 0; i < n; i++)
{
dis[i] = INF;
if(graph[i][i] != 0) return false;
for(int j = i+1 ; j < n; j++)
{
if(graph[i][j] != graph[j][i] || graph[i][j] == 0) return false;
}
}
memset(visit,0,sizeof(visit));
memset(prior,-1,sizeof(prior));
memset(f,0,sizeof(f));
int cent = n;
dis[0]=0;
while(cent--)//循环n次是由于要初始化
{
int min = -1;
for(int i = 0; i < n; i++)
{
if(!visit[i] && (min == -1 || dis[i] < dis[min]))
{
min = i;
}
}
for(int i = 0; i < n; i++)//在prim算法里面添加这层循环里面的内容算距离
{
if(visit[i])//必须是已经訪问过的点,才干算距离
{
f[i][min] = f[min][i] = f[i][prior[min]] + dis[min];
}
}
visit[min] = true;
for(int i = 0; i < n; i++)
{
if(dis[i] > graph[min][i] )
{
dis[i] = graph[min][i];
prior[i] = min;//记录前驱
}
}
}
for(int i = 0; i < n; i++)
{
for(int j = 0 ; j < n; j++)
{
if(f[i][j] != graph[i][j])
{
return false;
}
}
}
return true;
}
int main()
{
#ifdef xxz
freopen("in","r",stdin);
#endif // xxz
while(cin>>n)
{
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
{
cin>>graph[i][j];
}
if(check()) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return 0;
}
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