Problem Description

Harry Potter has a difficult homework. Given a rectangular table, consisting of n?×?m cells. Each cell of the table contains the integer. Harry knows how to use two spells: the first spell change the sign of the integers in the selected row, the second — in the selected column. Harry's task is to make non-negative the sum of the numbers in each row and each column using these spells.

Alone, the boy can not cope. Help the young magician!

Input

The first line contains two integers n and m (1?≤?n,??m?≤?100) — the number of rows and the number of columns.

Next n lines follow, each contains m integers: j-th integer in the i-th line is ai,?j (|ai,?j|?≤?100), the number in the i-th row and j-th column of the table.

The rows of the table numbered from 1 to n. The columns of the table numbered from 1 to m.

Output

In the first line print the number a — the number of required applications of the first spell. Next print a space-separated integers — the row numbers, you want to apply a spell. These row numbers must be distinct!

In the second line print the number b — the number of required applications of the second spell. Next print b space-separated integers — the column numbers, you want to apply a spell. These column numbers must be distinct!

If there are several solutions are allowed to print any of them.

Examples

Input

4 1
-1
-1
-1
-1

Output

4 1 2 3 4?
0

Input

2 4
-1 -1 -1 2
1 1 1 1

Output

1 1?
1 4?

題意：給出一個 n*m 的矩陣，其元素絕對值均小于 100，現在可以讓某一行或一列的所有數取反，要求構造一個解，使得每一行每一列的和都是非負數

思路：

首先統計矩陣每一行每一列的和，然后去枚舉行列的和，當某一行的值小于 0，那么將這一行直接翻轉，再對行列和進行相應處理，直到沒有負值

由于矩陣元素最多為 100 個，元素值范圍為 -100~100，因此，整個矩陣和的范圍為 [-10000,10000]，故而當某一行列的和為負時，將其進行翻轉后，和至少 +2，因此最多枚舉 100^4 即可令整個矩陣的行列和為正

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #include<bitset> #define EPS 1e-9 #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long const int MOD = 1E9+7; const int N = 5000+5; const int dx[] = {-1,1,0,0,-1,-1,1,1}; const int dy[] = {0,0,-1,1,-1,1,-1,1}; using namespace std;LL a[N][N]; LL sumRow[N],sumCol[N]; int row[N],col[N]; vector<int>resRow,resCol; int main() {int n,m;scanf("%d%d",&n,&m);for(int i=1; i<=n; i++) {for(int j=1; j<=m; j++) {scanf("%lld",&a[i][j]);sumRow[i]+=a[i][j];sumCol[j]+=a[i][j];}}while(true) {for(int i=1; i<=n; i++) {if(sumRow[i]<0) {sumRow[i]=0;for(int j=1; j<=m; j++) {a[i][j]=-a[i][j];sumRow[i]+=a[i][j];sumCol[j]=sumCol[j]-(-a[i][j])+a[i][j];}row[i]+=1;}}for(int j=1; j<=m; j++) {if(sumCol[j]<0) {sumCol[j]=0;for(int i=1; i<=n; i++) {a[i][j]=-a[i][j];sumCol[j]+=a[i][j];sumRow[i]=sumRow[i]-(-a[i][j])+a[i][j];}col[j]+=1;}}bool flag=false;for(int i=1; i<=n; i++) {if(sumRow[i]<0) {flag=true;}}for(int i=1; i<=m; i++) {if(sumCol[i]<0) {flag=true;}}if(!flag)break;}int num1=0,num2=0;for(int i=1; i<=n; i++) {if(row[i]%2==1) {resRow.push_back(i);num1++;}}for(int i=1; i<=m; i++) {if(col[i]%2==1) {resCol.push_back(i);num2++;}}printf("%d ",num1);for(int i=0; i<num1; i++)printf("%d ",resRow[i]);printf("\n");printf("%d ",num2);for(int i=0; i<num2; i++)printf("%d ",resCol[i]);printf("\n");return 0; }

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