zoukankan      html  css  js  c++  java
  • UVA 11464

    D

    Even Parity

    Input: Standard Input

    Output: Standard Output

    We have a grid of size N x N. Each cell of the grid initially contains a zero(0) or a one(1). 
    The parity of a cell is the number of 1s surrounding that cell. A cell is surrounded by at most 4 cells (top, bottom, left, right).

    Suppose we have a grid of size 4 x 4: 

    1

    0

    1

    0

    The parity of each cell would be

    1

    3

    1

    2

    1

    1

    1

    1

    2

    3

    3

    1

    0

    1

    0

    0

    2

    1

    2

    1

    0

    0

    0

    0

    0

    1

    0

    0

    For this problem, you have to change some of the 0s to 1s so that the parity of every cell becomes even. We are interested in the minimum number of transformations of 0 to 1 that is needed to achieve the desired requirement.

     
    Input

    The first line of input is an integer T (T<30) that indicates the number of test cases. Each case starts with a positive integer N(1≤N≤15). Each of the next N lines contain N integers (0/1) each. The integers are separated by a single space character.

    Output

    For each case, output the case number followed by the minimum number of transformations required. If it's impossible to achieve the desired result, then output -1 instead.

    Sample Input                             Output for Sample Input

    3
    3
    0 0 0
    0 0 0
    0 0 0
    3
    0 0 0
    1 0 0
    0 0 0
    3
    1 1 1
    1 1 1
    0 0 0
     

    Case 1: 0 
    Case 2: 3 
    Case 3: -1


    题意:给定n*n矩阵,可以把0变成1,求最少变幻次数使得每个位置的上下左右之和为偶数。

    思路:n为15,第一行状态最多2^15种,然后由前一行可以推出后一行,如此一来,时间复杂度为O(2^n) * (n^2).

    代码:

    #include <stdio.h>
    #include <string.h>
    #define min(a,b) (a)<(b)?(a):(b)
    #define INF 0x3f3f3f3f
    const int d[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    const int N = 15;
    int t, n, map[N][N], save[N][N];
    
    void init() {
        scanf("%d", &n);
        for (int i = 0; i < n; i ++)
    	for (int j = 0; j < n; j ++) {
    	    scanf("%d", &map[i][j]);
    	}
    }
    
    int Sum(int i, int j) {
        int sum = 0;
        for (int k = 0; k < 4; k ++) {
    	int xx = i + d[k][0];
    	int yy = j + d[k][1];
    	if (xx >= 0 && xx < n && yy >= 0 && yy < n) {
    	    sum += save[xx][yy];    
    	}
        }
        return sum;
    }
    
    int cal(int state) {
        int count = 0;
        memset(save, 0, sizeof(save));
        for (int i = 0; i < n; i ++)
    	for (int j = 0; j < n; j ++)
    	    save[i][j] = map[i][j];
        for (int i = n - 1; i >= 0; i --)
    	if (state >= (1<<i)) {
    	    if (!save[0][n - 1 - i])
    		count ++;
    	    save[0][n - 1 - i] = 1;
    	    state -= (1<<i);
    	}
        for (int i = 0; i < n - 1; i ++) {
    	for (int j = 0; j < n; j ++) {
    	    if (Sum(i, j) % 2) {
    		if (save[i + 1][j])
    		    return count = INF;
    		save[i + 1][j] = 1;
    		count ++;
    	    }
    	}
        }
        for (int j = 0; j < n; j ++)
    	if (Sum(n - 1, j) % 2) {
    	    count = INF;
    	    break;
    	}
        return count;
    }
    
    int judge(int state) {
        for (int i = 0; i < n; i ++)
    	if (map[0][i] == 1 && (state&(1<<i) == 0))
    	    return false;
        return true;
    }
    
    void solve() {
        int m = 1<<n, ans = INF;
        for (int i = 0; i < m; i ++) {
    	if (judge(i)) {
    	    ans = min(ans, cal(i));
    	}
        }
        if (ans == INF)
    	printf("-1
    ");
        else
    	printf("%d
    ", ans);
    }
    
    int main() {
        int cas = 0;
        scanf("%d", &t);
        while (t--) {
    	init();
    	printf("Case %d: ", ++cas);
    	solve();
        }
        return 0;
    }


  • 相关阅读:
    Matplotlib绘制漫威英雄战力图,带你飞起来!
    jupyter渲染网页的3种方式
    MySQL全文索引、联合索引、like查询、json查询速度大比拼
    进一步聊聊weight initialization
    深度学习基础(2)
    深度学习基础(1)
    SLAM的前世今生
    深度学习:识别图片中的电话号码(1)
    tf更新tensor/自定义层
    tf训练OTSU
  • 原文地址:https://www.cnblogs.com/riasky/p/3469153.html
Copyright © 2011-2022 走看看