wenbao与常见策略（大白书）

枚举

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=26&problem=2459&mosmsg=Submission+received+with+ID+18787313

 

01矩阵，使上下左右和为偶数

 

 1 #include <iostream>
 2 #include <string.h>
 3 using namespace std;
 4 #define ll long long
 5 #define REP(i, x, n) for(int i = x; i < n; i++)
 6 const int maxn = 17;
 7 const int INF = 1<<30;
 8 int n, a[maxn][maxn], b[maxn][maxn];
 9 int go(int x){
10     memset(b, 0, sizeof(b));
11     REP(i, 0, n){
12         if(x & (1<<i)) b[0][i] = 1;
13         else if(a[0][i] == 1) return INF;
14     }
15     REP(i, 1, n) REP(j, 0, n){
16         int sum = 0;
17         if(i > 1) sum += b[i-2][j];
18         if(j > 0) sum += b[i-1][j-1];
19         if(j < n-1) sum += b[i-1][j+1];
20         b[i][j] = sum%2;
21         if(b[i][j] == 0 && a[i][j] == 1) return INF;
22     }
23     int num = 0;
24     REP(i, 0, n) REP(j, 0, n){
25         if(a[i][j] != b[i][j]) num ++;
26     }
27     return num;
28 }
29 int main(){
30     int t;
31     scanf("%d", &t);
32     REP(i, 1, t+1){
33         int sum = INF;
34         scanf("%d", &n);
35         REP(j, 0, n) REP(k, 0, n) scanf("%d", &a[j][k]);
36         int x = 1 << n;
37         REP(j, 0, x) sum = min(sum, go(j));
38         printf("Case %d: %d\n", i, sum == INF ? -1 : sum);
39     }
40     return 0;
41 }

 

 

 

只有不断学习才能进步！

 

posted @ 2018-04-14 13:50  wenbao  阅读(107)  评论(0)  编辑  收藏  举报