【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 ;

 }

巴特西

【CF】270D Design Tutorial: Inverse the Problem

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