题意
请参考LRJ白书第一章例题7.
思路
大概是个带有机智成分的枚举。
为什么这么说，因为这道题的关键思路点就是发现只要知道第一行，就可以构造出整个数表，这样我们只要枚举第一行就可以解决问题，复杂度直接降低了一次幂。
代码
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#include <iostream> #include <algorithm> #include <queue> #include <vector> #include <cstdlib> #include <cstdio> #include <string> #include <cstring> #include <ctime> #include <iomanip> #include <cmath> #include <set> #include <stack> #include <cmath> #include <map> #include <complex> using namespace std; int n; int per[16]; int orig[16][16]; int now[16][16]; void genPermutation(int s){ int cur = n; while(cur){ if(s&1){ per[cur] = 1; }else{ per[cur] = 0; } cur--; s >>= 1; } } void genNow(){ memset(now,0,sizeof(now)); for(int i = 1; i <= n; i++){ now[1][i] = per[i]; } for(int i = 1; i < n; i++){ for(int j = 1; j <= n; j++){ int cnt = now[i-1][j] + now[i][j-1] + now[i][j+1]; if(cnt % 2){ now[i+1][j] = 1; }else{ now[i+1][j] = 0; } } } } int compare(){ int ret = 0; for(int i = 1; i <= n; i++){ for(int j = 1; j <= n; j++){ if(now[i][j] == 0 && orig[i][j] == 1){ return 0x3f3f3f3f; } if(now[i][j] != orig[i][j]){ ret++; } } } return ret; } int main(){ int caseCnt; scanf(" %d",&caseCnt); for(int d = 1; d <= caseCnt; d++){ scanf(" %d",&n); for(int i = 1; i <= n; i++){ for(int j = 1; j <= n; j++){ scanf(" %d",&orig[i][j]); } } int res = 0x3f3f3f3f; for(int i = 0; i < (1 << n); i++){ genPermutation(i); genNow(); res = min(res,compare()); } if(res == 0x3f3f3f3f){ res = -1; } printf("Case %d: %d\n",d,res); } return 0; } |