Code Jam 2009 Round2 Problem A. Crazy Rows —— 贪心

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.

Input

The first line of input gives the number of cases, TT 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.

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

Sample

Input 

Output 
 
3
2
10
11
3
001
100
010
4
1110
1100
1100
1000

Case #1: 0
Case #2: 2
Case #3: 4
本题的题意是给你一个n*n的01矩阵,让你交换最小的相邻两行的次数使得该矩阵成为下三角矩阵。

思路很简单,我去记录每一行的最后一个1的位置a[i],然后该行如果移动,移动至第i行最优。

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>

using namespace std;
#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
//#define mid ((l + r) >> 1)
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
#define bug(s) cout<<"#s = "<< s << endl;
const int MAXN = 100 ;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
#define mod 1000000007
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pi;
///#pragma comment(linker, "/STACK:102400000,102400000")
char G[MAXN][MAXN];
int a[MAXN];
int ans , n;
void solve()
{
    FOR(i , 0 , n)
    {
        a[i] = -1;
        FOR(j , 0 , n)if(G[i][j] == '1')
        {
            a[i] = j;
        }
    }
    ans = 0;
    FOR(i , 0 , n)
    {
        int pos = -1;
        FOR(j , i , n)
        {
            if(a[j] <= i)
            {
                pos = j;
                break;
            }
        }
        REP(j , pos , i)
        {
            swap(a[j] , a[j - 1]);
            ans++;
        }
    }
//    cout << ans << endl;
}
int main()
{
    //ios::sync_with_stdio(false);
    #ifdef Online_Judge
        freopen("A-large-practice.in","r",stdin);
        freopen("A-large-practice.out","w",stdout);
    #endif // Online_Judge
    int t ;
    cin >> t;
    FORR(kase , 1 , t)
    {
        scanf("%d" , &n);
        FOR(i , 0 , n)scanf("%s" , G[i]);
        ans = 0;
        solve();
        printf("Case #%d: %d\n" , kase , ans);
    }
    return 0;
}



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