Codeforces Round #270 Codeforces Round #270

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本文介绍了一种基于已知节点间距离矩阵逆向构造加权树的问题，并提供了一个具体的算法实现方案。首先通过最小生成树算法构建树结构，然后利用深度优先搜索验证构造的树是否与原始距离矩阵相匹配。

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).

Input

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.

Output

If there exists such a tree, output "YES", otherwise output "NO".

   Sample test(s)
 

     input
   
3
0 2 7
2 0 9
7 9 0

     output
   
YES

     input
   
3
1 2 7
2 0 9
7 9 0

     output
   
NO

     input
   
3
0 2 2
7 0 9
7 9 0

     output
   
NO

     input
   
3
0 1 1
1 0 1
1 1 0

     output
   
NO

     input
   
2
0 0
0 0

     output
   
NO

Note

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.

链接:http://codeforces.com/problemset/problem/472/D

题意：给个矩阵A（A(i,j)代表点i到点j的距离）判断满不满足一个树

思路:最小生成树后，DFS这颗树看u->v的距离是不是与矩阵A一致

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include<string>
#include <algorithm>
#define min(a,b)(a<b?a:b)
#define max(a,b)(a>b?a:b)
using namespace std;
typedef __int64 int64 ;
const int MAXN=2012;
int mp[MAXN][MAXN];
int n,k,m;
int head[MAXN];
int f[MAXN];
int p[MAXN];
struct HdNode
{
	int v,next,w;
}e[MAXN<<1];

struct Ed
{
	int u,v,w;
}edge[MAXN*MAXN>>1];

void add(int u,int v,int w)
{
	e[k].v=v;e[k].w=w;
	e[k].next=head[u];
	head[u]=k++;
}
void add1(int u,int v,int w)
{
	edge[m].v=v;edge[m].w=w;
	edge[m++].u=u;
}

bool cmp(Ed x,Ed y) {return x.w<y.w;}

int find(int x) {return x==p[x]?x:p[x]=find(p[x]);}

bool dfs(int s,int u,int64 d)
{
	bool y=1;
	if((int64)mp[s][u]!=d) return false;
	f[u]=true;
	for(int i=head[u];i!=-1;i=e[i].next)
	{
		if(f[e[i].v])continue;
		y=y && dfs(s,e[i].v,(int64)e[i].w+d);
	}
	return y;
}

bool solve()
{
	int i,j;

	for(i=1;i<=n;i++) p[i]=i;
	memset(head,-1,sizeof(head));
	sort(edge,edge+m,cmp);

	int cnt=0;
	for(i=0;i<m;i++)
	{
		int u=find(edge[i].u),v=find(edge[i].v);
		if(u!=v)
		{
			p[u]=v;	cnt++;
			add(edge[i].u,edge[i].v,edge[i].w);
			add(edge[i].v,edge[i].u,edge[i].w);
			if(cnt==n-1) break;
		}
	}

	for(i=1;i<=n;i++)
	{
		memset(f,0,sizeof(f));
		f[i]=true;	
		if(!dfs(i,i,0)) return false;
	}
	return true;
}


int Scan()
{
	int res = 0 , ch;
	while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
	{
		if( ch == EOF )  return 1 << 30 ;
	}
	res = ch - '0' ;
	while( ( ch = getchar() ) >= '0' && ch <= '9' )
		res = res * 10 + ( ch - '0' ) ;
	return res ;
}

int main()
{
	int i,j,yes=0;
	n=Scan();
	//scanf("%d",&n);
	m=k=0;
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=n;j++)
		{
			mp[i][j]=Scan();
		//	scanf("%d",&mp[i][j]);
			if(i<j) add1(i,j,mp[i][j]);
			else if(i==j && mp[i][j])yes=1;
			else if(i>j && (mp[i][j]!=mp[j][i] || mp[i][j]==0))yes=1;
		}
	}

	if(yes)	puts("NO");
	else if(n==1 || solve()) puts("YES");
	else puts("NO");
	//main();
	return 0;
}