CF274D：Lovely Matrix（拓扑排序 & 缩点）

D. Lovely Matrix time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

Lenny had an n × m matrix of positive integers. He loved the matrix so much, because each row of the matrix was sorted in non-decreasing order. For the same reason he calls such matrices of integers lovely.

One day when Lenny was at school his little brother was playing with Lenny's matrix in his room. He erased some of the entries of the matrix and changed the order of some of its columns. When Lenny got back home he was very upset. Now Lenny wants to recover his matrix.

Help him to find an order for the columns of the matrix so that it's possible to fill in the erased entries of the matrix to achieve a lovely matrix again. Note, that you can fill the erased entries of the matrix with any integers.

Input

The first line of the input contains two positive integers n and m (1 ≤ n·m ≤ 105). Each of the next n lines contains m space-separated integers representing the matrix. An integer -1 shows an erased entry of the matrix. All other integers (each of them is between 0 and 109inclusive) represent filled entries.

Output

If there exists no possible reordering of the columns print -1. Otherwise the output should contain m integers p1, p2, ..., pm showing the sought permutation of columns. So, the first column of the lovely matrix will be p1-th column of the initial matrix, the second column of the lovely matrix will be p2-th column of the initial matrix and so on.

Examples input 3 3 1 -1 -1 1 2 1 2 -1 1 output 3 1 2 input 2 3 1 2 2 2 5 4 output 1 3 2 input 2 3 1 2 3 3 2 1 output -1 题意：一个N*M的数列，要求每一行的数字都是非递减的，可以交换某些列达到这一要求，-1代表未知数，输出重新安排后的列，不行输出-1。

思路：拓扑排序，若每一行都逐个建边需要O(n*m*m)，那么先排序，通过定义“冗余点”进行统一建边，就变成O(n*m*log(m))了。

#include <bits/stdc++.h> # define pb push_back using namespace std; const int maxn = 2e5+300; int flag=1, n, m, a[maxn], in[maxn], q[maxn]; int ans[maxn], cnt=0; vector<int>v[maxn]; struct node{int val, id;}e[maxn]; bool cmp(node x, node y){return x.val < y.val;} void add(int x, int y) { v[x].pb(y); ++in[y]; } void toposort() { int l=0, r=0, icount=0; for(int i=1; i<cnt+m; ++i) if(in[i] == 0) q[r++] = i; while(l<r) { int u = q[l++]; if(u <= m) ans[icount++] = u; for(auto j : v[u]) if(--in[j] == 0) q[r++] = j; } if(icount < m) flag = 0; } int main() { scanf("%d%d",&n,&m); for(int i=1; i<=n; ++i) { for(int j=1; j<=m; ++j) scanf("%d",&e[j].val),e[j].id=j; sort(e+1, e+1+m, cmp); for(int j=1; j<=m; ++j) { if(e[j].val == -1) continue; if(j==1||e[j].val != e[j-1].val) ++cnt; add(e[j].id, cnt+m+1); add(cnt+m, e[j].id); } ++cnt; } toposort(); if(!flag) puts("-1"); else for(int i=0; i<m; ++i) printf("%d ",ans[i]); return 0; }