D - Fliptile——POJ
来源:互联网 发布:5g网络哪些手机支持 编辑:程序博客网 时间:2024/06/06 17:20
Description
Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for cows in which they manipulate an M × N grid (1 ≤ M ≤ 15; 1 ≤ N ≤ 15) of square tiles, each of which is colored black on one side and white on the other side.
As one would guess, when a single white tile is flipped, it changes to black; when a single black tile is flipped, it changes to white. The cows are rewarded when they flip the tiles so that each tile has the white side face up. However, the cows have rather large hooves and when they try to flip a certain tile, they also flip all the adjacent tiles (tiles that share a full edge with the flipped tile). Since the flips are tiring, the cows want to minimize the number of flips they have to make.
Help the cows determine the minimum number of flips required, and the locations to flip to achieve that minimum. If there are multiple ways to achieve the task with the minimum amount of flips, return the one with the least lexicographical ordering in the output when considered as a string. If the task is impossible, print one line with the word "IMPOSSIBLE".
Input
Lines 2.. M+1: Line i+1 describes the colors (left to right) of row i of the grid with N space-separated integers which are 1 for black and 0 for white
Output
Sample Input
4 41 0 0 10 1 1 00 1 1 01 0 0 1
Sample Output
0 0 0 01 0 0 11 0 0 10 0 0 0
自写:
#include<stdio.h>#include<string.h>int n,m;int inf=999999,res;int flag[20][20];int mid[20][20];int end[20][20];int dir[5][2]={{0,0},{1,0},{-1,0},{0,1},{0,-1}};int get(int x,int y);int cal();int main(){int i,j;while(scanf("%d %d",&m,&n)!=EOF){for(i=1;i<=m;i++)//录入 {for(j=1;j<=n;j++){scanf("%d",&flag[i][j]);}}res=inf;f<span style="font-family: Arial, Helvetica, sans-serif;">or(i=0;i<(1<<n);i++)</span><pre name="code" class="cpp">{memset(mid,0,sizeof(mid));for(j=0;j<n;j++)//字典序,先确定第一个 {mid[1][n-j]=i>>j&1;}int num=cal();//从第二行开始向下遍历 if(num>0&&num<res)//更新最小值 {res=num;memcpy(end,mid,sizeof(mid));}}
if(res==inf)printf("IMPOSSIBLE\n");else{for(i=1;i<=m;i++){for(j=1;j<=n;j++){if(j-1) printf(" ");printf("%d",end[i][j]);}printf("\n");}}}return 0;}int get(int x,int y)//判断其目前状态,是正还是负 {int i,xx,yy,sum=flag[x][y];for(i=0;i<5;i++){xx=x+dir[i][0];yy=y+dir[i][1];if(xx>=0&&xx<=m&&yy>=0&&yy<=n){sum+=mid[xx][yy];}}return sum%2;}int cal(){int i,j,sum=0;for(i=2;i<=m;i++)for(j=1;j<=n;j++){if(get(i-1,j)!=0)//跟据同一列的上一个行的数,来判断其是否 要反转 mid[i][j]=1;}for(j=1;j<=n;j++)if(get(m,j)!=0)//判断最后一行,若最后一行还需要反转,则其无法反转 return -1;for(i=1;i<=m;i++)for(j=1;j<=n;j++){sum+=mid[i][j];}return sum;}
知识点:
利用二进制数字典序:
<span style="white-space:pre"></span>for(i=0;i<(1<<n);i++){for(j=0;j<n;j++)//字典序,先确定第一个 {mid[1][n-j]=i>>j&1;}}
知识点2:
从头到尾,覆盖!
参考:
#include<iostream>#include<cstring>#include<cstdio>#include<climits>using namespace std;const int int_max = INT_MAX;const int dirx[5] = { 1, -1, 0, 0, 0 };const int diry[5] = { 0, 0, 0, 1, -1 };int tile[20][20]; //黑白标志 int flip[20][20];//int opt[20][20];//int n, m;//??(x,y)???int get(int x, int y){int c = tile[x][y];for (int i = 0; i < 5; i++){int xx = x + dirx[i],yy = y + diry[i];if (xx >= 0 && xx < m && yy >= 0 && yy < n)c += flip[xx][yy];}return c % 2;}//???????????????,????-1int calc(){//???????????for (int i = 1; i < m;i++)//第二行 for (int j = 0; j < n; j++){if (get(i - 1, j) != 0)flip[i][j] = 1;}//判断最后是否为全白 for (int j = 0; j < n;j++)if (get(m - 1, j) != 0)return -1;//统计反转次数 int ret = 0;for (int i = 0; i < m; i++)for (int j = 0; j < n; j++)ret += flip[i][j];return ret;}int main(){while (scanf("%d%d", &m, &n) != EOF){for (int i = 0; i < m;i++)for (int j = 0; j < n; j++)scanf("%d", &tile[i][j]);int res = int_max;//字典序 for (int i = 0; i < (1 << n); i++){memset(flip, 0, sizeof(flip));for(int j = 0; j < n; j++)flip[0][n - j - 1] = i >> j & 1;int num = calc();if (num > 0 && num < res){res = num;memcpy(opt,flip,sizeof(flip));}}if (res == int_max) printf("IMPOSSIBLE\n");else{for (int i = 0; i < m; i++){for (int j = 0; j < n; j++){if (j) printf(" ");printf("%d", opt[i][j]);}printf("\n");}}}}
- D - Fliptile——POJ
- POJ 3279 Fliptile——开关问题
- Fliptile POJ
- Fliptile POJ
- Fliptile POJ
- Fliptile POJ
- POJ 3279 Fliptile (暴力枚举)(D)
- poj 3279 Fliptile—二进制暴力的运用(开关问题)
- POJ-3279-Fliptile
- POJ 3279 Fliptile
- poj 3279 Fliptile
- POJ 3279 Fliptile
- poj 3279 Fliptile
- POJ 3279 Fliptile
- poj 3279 Fliptile
- Poj 3279 Fliptile 【枚举】
- poj 3297 Fliptile 深搜
- POJ 3279 Fliptile
- 关于自定义listview,整合下拉刷新上拉加载功能,以及item侧滑功能,并且解决滑动冲突
- 我的LaTeX秘籍(不断更新中)
- ajax+jsonp+php 实现跨域请求
- Android 一个改善的okHttp封装库
- 性能调优攻略
- D - Fliptile——POJ
- php命名空间
- http web服务器
- VPN的使用
- 链表——partition list
- Java JTable与滚动条JScollpane结合使用不能出现很想滚动条的解决办法
- Android自动回放工具——RERAN
- 【小笔记】修改状态条文字的颜色为白色的
- 开源数据集