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  • Crazy Rows

    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

    代码:

     1 int n;
     2 int mp[MAX][MAX];  //矩阵
     3 
     4 int a[MAX];   //表示第i行最后出现1的位置
     5 
     6 void solve()
     7 {
     8     int ans=0;
     9     for(int i=0; i<n; i++){
    10         a[i]=-1;
    11         for(int j=0; j<n; j++){
    12             if(mp[i][j]==1)
    13                 a[i]=j;
    14         }
    15     }
    16     for(int i=0; i<n; i++){
    17         int pos=-1;
    18         for(int j=i; j<n; j++){
    19             if(a[j]<=i){
    20                 pos=j;
    21                 break;
    22             }
    23         }
    24 
    25         for(int j=pos; j>i; j--){
    26             swap(a[j],a[j-i]);
    27             ans++;
    28         }
    29     }
    30     printf("%d",&ans);
    31 }
    View Code
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  • 原文地址:https://www.cnblogs.com/wangmengmeng/p/5323218.html
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