Codeforces 472(#270 ) D Design Tutorial: Inverse the Problem

题意：给你一棵数的距离矩阵，问你这个矩阵是否合法

解题思路：利用树的性质进行prime 和连边，产生最小生成树，最后看最小生成树是否和给出的一致就行

解题代码：

  1 // File Name: d.cpp
  2 // Author: darkdream
  3 // Created Time: 2014年09月29日 星期一 00时45分45秒
  4
  5 #include<vector>
  6 #include<list>
  7 #include<map>
  8 #include<set>
  9 #include<deque>
 10 #include<stack>
 11 #include<bitset>
 12 #include<algorithm>
 13 #include<functional>
 14 #include<numeric>
 15 #include<utility>
 16 #include<sstream>
 17 #include<iostream>
 18 #include<iomanip>
 19 #include<cstdio>
 20 #include<cmath>
 21 #include<cstdlib>
 22 #include<cstring>
 23 #include<ctime>
 24 #define LL long long
 25 #define maxn 2010
 26 using namespace std;
 27 int n ;
 28 int mi[maxn];
 29 vector <int> mp[2010];
 30 int nmp[maxn][maxn];
 31 int pre[maxn];
 32 bool vis[maxn];
 33
 34 long long  dep[maxn];
 35 void dfs(int la,int zla,int dis)
 36 {
 37    dep[la] = dis;
 38    int len = mp[la].size();
 39    for(int i = 0 ;i < len ;i ++)
 40    {
 41        int ne = mp[la][i];
 42        if(ne != zla)
 43        {
 44           dfs(ne,la,dis + nmp[la][ne]);
 45        }
 46    }
 47 }
 48 void prime()
 49 {
 50     for(int i = 1;i <= n;i ++){
 51        dep[i] = 2e9;
 52        vis[i] = 0 ;
 53     }
 54     dep[1] = 0 ;
 55     while(1)
 56     {
 57        int v = -1;
 58        for(int i = 1;i <= n;i ++)
 59        {
 60           if(!vis[i] && (v == -1 || dep[i] < dep[v]))
 61               v = i ;
 62        }
 63        if(v == -1)
 64            break;
 65        vis[v] = 1;
 66        for(int i = 1;i <= n;i ++)
 67        {
 68            if(!vis[i] && nmp[v][i] < dep[i])
 69            {
 70               dep[i] = nmp[pre[i] = v][i];
 71            }
 72        }
 73     }
 74 }
 75 int main(){
 76      scanf("%d",&n);
 77      int sum = 0 ;
 78      int ok  = 0 ;
 79      for(int i = 1;i <= n;i ++)
 80      {
 81         for(int j = 1;j <= n;j ++)
 82         {
 83             scanf("%d",&nmp[i][j]);
 84         }
 85      }
 86      for(int i = 1;i <= n;i ++)
 87      {
 88         for(int j = 1;j <= n;j ++)
 89         {
 90            if(i!= j && nmp[i][j] == 0 )
 91                ok = 1;
 92            if(nmp[i][j] != nmp[j][i])
 93                ok = 1;
 94         }
 95      }
 96      if(ok)
 97      {
 98        puts("NO");
 99        return 0 ;
100      }
101      prime();
102      for(int i = 2;i <= n;i ++)
103      {
104        mp[i].push_back(pre[i]);
105        mp[pre[i]].push_back(i);
106        //printf("%d %d\n",i,pre[i]);
107      }
108      for(int i = 1;i <= n;i ++)
109      {
110         dfs(i,0,0);
111     /*    for(int j = 1;j <= n;j ++)
112             printf("%lld ",dep[j]);
113         printf("\n");*/
114         for(int j = 1; j <= n;j ++)
115         {
116            if(dep[j] != nmp[i][j])
117            {
118
119               puts("NO");
120               return 0;
121            }
122         }
123      }
124      puts("YES");
125 return 0;
126 }