【CF】270D Design Tutorial: Inverse the Problem

题意异常的简单。就是给定一个邻接矩阵，让你判定是否为树。
算法1：O(n^3)。思路就是找到树边，原理是LCA。判断树边的数目是否为n-1。39-th个数据T了，自己测试2000跑到4s。
算法2：O(n^2)。思考由图如何得到树，显然MST是可行的。因此，题目变为直接找到MST。然后通过树边构建目标矩阵。判定矩阵是否相等。

 /* 472D */
 #include <iostream>
 #include <string>
 #include <map>
 #include <queue>
 #include <set>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cassert>
 #include <functional>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,1024000")

 #define mpii            map<int,int>
 #define vi                vector<int>
 #define pii                pair<int,int>
 #define vpii            vector<pair<int,int> >
 #define rep(i, a, n)     for (int i=a;i<n;++i)
 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 #define pb                 push_back
 #define mp                 make_pair
 #define fir                first
 #define sec                second
 #define all(x)             (x).begin(),(x).end()
 #define SZ(x)             ((int)(x).size())
 #define lson            l, mid, rt<<1
 #define rson            mid+1, r, rt<<1|1

 const int maxn = ;
 const int INF = 0x3f3f3f3f;
 int M[maxn][maxn];
 int T[maxn][maxn];
 int dist[maxn];
 int fa[maxn];
 bool visit[maxn];
 int n;

 void kruskal() {
     memset(visit, false, sizeof(visit));
     memset(dist, 0x3f, sizeof(dist));
     int v, mn;
     int i, j, k;

     dist[] = ;
     for (i=; i<n; ++i) {
         mn = INF;
         for (j=; j<n; ++j) {
             if (!visit[j] && dist[j]<mn) {
                 mn = dist[j];
                 v = j;
             }
         }

         assert(mn < INF);

         for (j=; j<n; ++j) {
             if (visit[j]) {
                 T[j][v] = T[v][j] = T[j][fa[v]] + mn;
             }
         }
         visit[v] = true;

         for (j=; j<n; ++j) {
             if (M[v][j] < dist[j]) {
                 dist[j] = M[v][j];
                 fa[j] = v;
             }
         }
     }
 }

 int main() {
     ios::sync_with_stdio(false);
     #ifndef ONLINE_JUDGE
         freopen("data.in", "r", stdin);
         freopen("data.out", "w", stdout);
     #endif

     bool flag = true;

     scanf("%d", &n);

     rep(i, , n)
         rep(j, , n)
             scanf("%d", &M[i][j]);

     rep(i, , n) {
         if (M[i][i] != ) {
             flag = false;
             goto _output;
         }
         rep(j, i+, n) {
             if (M[i][j]!=M[j][i] || M[i][j]==) {
                 flag = false;
                 goto _output;
             }
         }
     }

     kruskal();

     rep(i, , n) {
         rep(j, i+, n) {
             if (M[i][j] != T[i][j]) {
                 flag = false;
                 goto _output;
             }
         }
     }

     _output:
     puts(flag ? "YES":"NO");

     #ifndef ONLINE_JUDGE
         printf("time = %d.\n", (int)clock());
     #endif

     return ;
 }