【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 = 2005;
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[0] = 0;
    for (i=0; i<n; ++i) {
        mn = INF;
        for (j=0; j<n; ++j) {
            if (!visit[j] && dist[j]<mn) {
                mn = dist[j];
                v = j;
            }
        }
        
        assert(mn < INF);
        
        for (j=0; j<n; ++j) {
            if (visit[j]) {
                T[j][v] = T[v][j] = T[j][fa[v]] + mn;
            }
        }
        visit[v] = true;
        
        for (j=0; 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, 0, n)
        rep(j, 0, n)
            scanf("%d", &M[i][j]);
            
    rep(i, 0, n) {
        if (M[i][i] != 0) {
            flag = false;
            goto _output;
        }
        rep(j, i+1, n) {
            if (M[i][j]!=M[j][i] || M[i][j]==0) {
                flag = false;
                goto _output;
            }
        }
    }
    
    kruskal();
    
    rep(i, 0, n) {
        rep(j, i+1, n) {
            if (M[i][j] != T[i][j]) {
                flag = false;
                goto _output;
            }
        }
    }
    
    _output:
    puts(flag ? "YES":"NO");
    
    #ifndef ONLINE_JUDGE
        printf("time = %d.
", (int)clock());
    #endif
    
    return 0;
}