Codeforces 472D. Design Tutorial: Inverse the Problem

D. Design Tutorial: Inverse the Problem
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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”.

Examples
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.
將給出的圖求出最小生成樹，然後在將最小生成樹中任意兩點距離與原矩陣比較即可。
證明並不是很難，與鬆弛操作差不多。

#include<stdio.h>
#include<string>
#include<algorithm>
#include<string.h>
#include<iostream>
#include<queue>
using namespace std;
#define F(x,a,b) for (int x=a;x<=b;x++)
#define me(x) memset(x,0,sizeof(x))
#define mme(x) memset(x,-1,sizeof(x))
#define Ea(x) _f(f[EE[x].a])
#define Eb(x) _f(f[EE[x].b])
#define reset F(i,1,n) f[i]=i;
#define ll long long
ll fr,head[2005],vis[2005],ma[2005][2005],f[2005],a[2005][2005];
struct p{ll a,b,v;}EE[2000005];
struct q{ll node,next,v;}E[200005];
bool cmp(p aa,p bb){return aa.v<bb.v;}
int _f(int x) {return f[x]=f[x]==x?x:_f(f[x]);}
void insertt(int x,int y,ll val){E[fr]={y,head[x],val};head[x]=fr++;}
void dfs(int st,int stand,int pr)
{
   vis[st]=1;
   for (int i=head[st];i!=-1;i=E[i].next)
   {
      if (!vis[E[i].node])
    {
       ma[stand][E[i].node]=pr+E[i].v;
       dfs(E[i].node,stand,ma[stand][E[i].node]);
    }
   }
}
int main()
{
    int n;int cnt=0; fr=0;int coutt=0;int flag=1;
    scanf("%d",&n);
    F(i,1,n)F(j,1,n){scanf("%d",&a[i][j]);if (a[i][j]&&i<j){flag=0;EE[++cnt].a=i;EE[cnt].b=j;EE[cnt].v=a[i][j];}}
    if (n==1&&!a[1][1]) {printf("YES\n");return 0;}
    if (flag) {printf("NO\n");return 0;}
    F(i,1,n)F(j,1,n){if (a[i][j]!=a[j][i]||(a[i][j]&&i==j)||(i!=j&&a[i][j]==0)){printf("NO\n");return 0;}}
    sort(EE+1,EE+cnt+1,cmp); reset mme(head);me(vis);
    F(i,1,cnt)
    {
        if (Ea(i)!=Eb(i))
        {
            f[Ea(i)]=Eb(i);
            coutt++;
            insertt(EE[i].a,EE[i].b,EE[i].v);insertt(EE[i].b,EE[i].a,EE[i].v);
        }
        if (coutt==n-1) break;
    }
    F(i,1,n){me(vis);dfs(i,i,0);}
    F(i,1,n)F(j,1,n){if (a[i][j]!=ma[i][j]){printf("NO\n");return 0;}}
    printf("YES\n");


}