[Usaco2007 Open]Fliptile 翻格子游戏

题目描述

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 x 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".

约翰知道，那些高智力又快乐的奶牛产奶量特别高．所以他做了一个翻瓦片的益智游戏来娱乐奶牛．在一个M×N(1≤M，N≤15)的骨架上，每一个格子里都有一个可以翻转的瓦片．瓦片的一面是黑色的，而另一面是白色的．对一个瓦片进行翻转，可以使黑变白，也可以使白变黑．然而，奶牛们的蹄子是如此的巨大而且笨拙，所以她们翻转一个瓦片的时候，与之有公共边的相邻瓦片也都被翻转了．那么，这些奶牛们最少需要多少次翻转，使所有的瓦片都变成白面向上呢？如杲可以做到，输出字典序最小的结果（将结果当成字符串处理）．如果不能做到，输出“IMPOSSIBLE”．

输入

* Line 1: Two space-separated integers: M and N

* 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

第1行输入M和N，之后M行N列，输入游戏开始时的瓦片状态．0表示白面向上，1表示黑面向上．

输出

* Lines 1..M: Each line contains N space-separated integers, each specifying how many times to flip that particular location.

输出M行，每行N个用空格隔开的整数，表示对应的格子进行了多少次翻转．

样例输入

样例输出

0 0 0 0
1 0 0 1
1 0 0 1
0 0 0 0

OUTPUT DETAILS:

After flipping at row 2 column 1, the board will look like:
0 0 0 1
1 0 1 0
1 1 1 0
1 0 0 1

After flipping at row 2 column 4, the board will look like:
0 0 0 0
1 0 0 1
1 1 1 1
1 0 0 1

After flipping at row 3 column 1, the board will look like:
0 0 0 0
0 0 0 1
0 0 1 1
0 0 0 1

After flipping at row 3 column 4, the board will look like:
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0

Another solution might be:
0 1 1 0
0 0 0 0
0 0 0 0
0 1 1 0
but this solution is lexicographically higher than the solution above.
solution:
　　　这题上来大暴搜，搜到无能为力QAQ,5*5的方格就跳不出来了，然后我们再想，先不管是否合法，只要第一行状态确定，下面的状态其实就可以由上一行退出来，1代表这点翻，0代表不翻，只要一个点状态和它四周加和再加上原来的矩阵的中的黑色或白色所代表的数字的和为偶数，则当前点的最终颜色为白色。
　　　  而且数据范围很小，一看就是状压dp，先枚举第一行状态，最后判断是否合法且为最优解即可。

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 using namespace std;
  6 int n,m,mx;
  7 int singl[17][17],state[2],change_num[17][17];
  8 void change() {
  9     memset(change_num,0,sizeof(change_num));
 10     for(int i=1; i<=n; i++) {
 11         if((1<<(n-i))&state[1]) {
 12             change_num[1][i]=1;
 13         }
 14     }
 15 }
 16 int sum,ans=0x7ffffff;
 17 bool check(int a[][17]) {
 18     int s[17][17];
 19     for(int i=1; i<=m; i++) {
 20         for(int j=1; j<=n; j++) {
 21             s[i][j]=singl[i][j];
 22         }
 23     }
 24     for(int i=1; i<=m; i++) {
 25         for(int j=1; j<=n; j++) {
 26             if(a[i][j]) {
 27                 s[i-1][j]^=1;
 28                 s[i+1][j]^=1;
 29                 s[i][j-1]^=1;
 30                 s[i][j+1]^=1;
 31                 s[i][j]^=1;
 32             }
 33         }
 34     }
 35     for(int i=1; i<=m; i++) {
 36         for(int j=1; j<=n; j++) {
 37             if(s[i][j]) {
 38                 return false;
 39             }
 40         }
 41     }
 42     return true;
 43 }
 44  
 45 void tra(int x) {
 46     state[1]=x;
 47     change();
 48     for(int i=1; i<m; i++) {
 49         for(int j=1; j<=n; j++) {
 50             if(j==1) {
 51                 int ji=(change_num[i-1][j]+singl[i][j]+change_num[i][j+1])&1;
 52                 if(ji&&change_num[i][j]) {
 53                     change_num[i+1][j]=0;
 54                 }
 55                 if(ji&&!change_num[i][j]) {
 56                     change_num[i+1][j]=1;
 57                 }
 58                 if(!ji&&change_num[i][j]) {
 59                     change_num[i+1][j]=1;
 60                 }
 61                 if(!ji&&!change_num[i][j]) {
 62                     change_num[i+1][j]=0;
 63                 }
 64                 continue;
 65             }
 66             if(j==n) {
 67                 int ji=(change_num[i-1][j]+singl[i][j]+change_num[i][j-1])&1;
 68                 if(ji&&change_num[i][j]) {
 69                     change_num[i+1][j]=0;
 70                 }
 71                 if(ji&&!change_num[i][j]) {
 72                     change_num[i+1][j]=1;
 73                 }
 74                 if(!ji&&change_num[i][j]) {
 75                     change_num[i+1][j]=1;
 76                 }
 77                 if(!ji&&!change_num[i][j]) {
 78                     change_num[i+1][j]=0;
 79                 }
 80                 continue;
 81             }
 82             int ji=(change_num[i-1][j]+singl[i][j]+change_num[i][j-1]+change_num[i][j+1])&1;
 83             if(ji&&change_num[i][j]) {
 84                 change_num[i+1][j]=0;
 85             }
 86             if(ji&&!change_num[i][j]) {
 87                 change_num[i+1][j]=1;
 88             }
 89             if(!ji&&change_num[i][j]) {
 90                 change_num[i+1][j]=1;
 91             }
 92             if(!ji&&!change_num[i][j]) {
 93                 change_num[i+1][j]=0;
 94             }
 95         }
 96     }
 97 }
 98 int cun[17][17];
 99 bool pp() {
100     for(int i=1; i<=m; i++) {
101         for(int j=1; j<=n; j++) {
102             if(cun[i][j]<change_num[i][j]) {
103                 return false;
104             }
105             if(cun[i][j]>change_num[i][j]) {
106                 return true;
107             }
108         }
109     }
110     return false;
111 }
112 int main() {
113     memset(cun,0x3f,sizeof(cun));
114     scanf("%d%d",&m,&n);
115     for(int i=1; i<=m; i++) {
116         for(int j=1; j<=n; j++) {
117             scanf("%d",&singl[i][j]);
118         }
119     }
120     mx=(1<<n);
121     bool pd=0;
122     for(int i=0; i<mx; i++) {
123         tra(i);
124         if(check(change_num)) {
125             sum=0;
126             pd=1;
127             for(int i=1; i<=m; i++) {
128                 for(int j=1; j<=n; j++) {
129                     if(change_num[i][j]) {
130                         ++sum;
131                     }
132                 }
133             }
134             if(sum==ans&&pp()) {
135                 for(int i=1; i<=m; i++) {
136                     for(int j=1; j<=n; j++) {
137                         cun[i][j]=change_num[i][j];
138                     }
139                 }
140             }
141             if(sum<ans) {
142                 ans=sum;
143                 for(int i=1; i<=m; i++) {
144                     for(int j=1; j<=n; j++) {
145                         cun[i][j]=change_num[i][j];
146                     }
147                 }
148             }
149         }
150     }
151     if(!pd) {
152         printf("IMPOSSIBLE");
153     } else {
154         for(int i=1; i<=m; i++) {
155             for(int j=1; j<=n; j++) {
156                 printf("%d ",cun[i][j]);
157             }
158             printf("\n");
159         }
160     }
161     return 0;
162 }

posted @ 2017-08-03 07:13 Forever_goodboy 阅读(218) 评论(0) 编辑收藏举报

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Forever_goodboy

守得住繁华耐得住寂寞抵得住诱惑你想要的岁月都会给

[Usaco2007 Open]Fliptile 翻格子游戏

题目描述

输入

输出

样例输入

样例输出

公告

Forever_goodboy

守得住繁华 耐得住寂寞 抵得住诱惑 你想要的 岁月都会给

[Usaco2007 Open]Fliptile 翻格子游戏

题目描述

输入

输出

样例输入

样例输出

公告

守得住繁华耐得住寂寞抵得住诱惑你想要的岁月都会给