Crazy Rows(2009 Round2 A)

Problem

You are given an N x N matrix with 0 and 1 values. You can swap any two adjacent rows of the matrix.

Your goal is to have all the 1 values in the matrix below or on the main diagonal. That is, for each X where 1 ≤ X ≤ N, there must be no 1 values in row X that are to the right of column X.

Return the minimum number of row swaps you need to achieve the goal.

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Input

The first line of input gives the number of cases, T. T test cases follow.
The first line of each test case has one integer, N. Each of the next N lines contains Ncharacters. Each character is either 0 or 1.

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Output

For each test case, output

Case #X: K

where X is the test case number, starting from 1, and K is the minimum number of row swaps needed to have all the 1 values in the matrix below or on the main diagonal.
You are guaranteed that there is a solution for each test case.

Limits

1 ≤ T ≤ 60

Small dataset

1 ≤ N ≤ 8

Large dataset

1 ≤ N ≤ 40

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Sample

Input

3

2
10
11

3

001
100
010

4

1110
1100
1100
1000

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Output

Case #1: 0
Case #2: 2
Case #3: 4

---

#include<bits/stdc++.h>
using namespace std;
#define MAX_N 50
int m[MAX_N][MAX_N];
int a[MAX_N],n;
void solve(){
	int res=0;
	for(int i=0;i<n;i++){
		a[i]=-1;//如果第i行不含 1 
		for(int j=0;j<n;j++){
			if(m[i][j]==1)
				a[i]=j;//得到每行最後一個 1 的位置 
		}
	}
	for(int i=0;i<n;i++){
		int pos=-1;//要移到第i行的行 
		for(int j=i;j<n;j++){
			if(a[j]<=i){
				pos=j;
				break;
			}
		}
		//完成轉換
		for(int j=pos;j>i;j--){
			swap(a[j],a[j-1]);
			res++;
		} 
	}	
	printf("%d\n",res);
//	for(int i=0;i<n;i++){//輸出查看 
//		for(int j=0;j<n;j++){
//			printf("%d ",m[i][j]);
//		}
//		printf("\n");
//	}
}
int main(){
	while(scanf("%d",&n)!=EOF){
		memset(m,0,sizeof(m));
		memset(a,0,sizeof(a));
		for(int i=0;i<n;i++){
			for(int j=0;j<n;j++){
				scanf("%d",&m[i][j]);
			}
		}
		solve();
		printf("\n"); 
	}
	return 0;
}

題解：如果採用嘗試所有 N! 的種交換方案，當 N 大於40的時候時間複雜度就特別大，所以不可行，本題從第一行開始一行一行的尋找每行的最後一個 1 的位置，然後交換直到最後一行。