【UVA】11464-Even Parity（二进制枚举子集）

枚举第一行的所有可能情况，之后根据上面行计算下面行（判断是否冲突），获得最终结果。

14058243

11464

Even Parity

Accepted

C++

0.275

2014-08-18 05:14:15

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<list>
#include<cmath>
#include<string>
#include<sstream>
#include<ctime>
using namespace std;
#define _PI acos(-1.0)
#define esp 1e-9
#define INF 1 << 30
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pill;
/*===========================================
===============KinderRiven===================
===========================================*/
#define MAXD 15 + 5
int n;
int Map[MAXD][MAXD];
int solve(int u){
    int MAP[MAXD][MAXD];
    int ans = 0;
    memcpy(MAP,Map,sizeof(MAP));
    for(int i = 0 ; i < n ; i++){
         if(!((u >> i) & 1))
            continue;
         else{
            if(MAP[0][i] == 1);
            else{
                MAP[0][i] = 1;
                ans ++;
            }
         }
    }
    for(int i = 1 ; i < n ; i ++)
        for(int j = 0 ; j < n ; j ++){
             int sum = 0;
             if(i > 1) sum += MAP[i - 2][j];
             if(j > 0) sum += MAP[i - 1][j - 1];
             if(j < n - 1)
             sum += MAP[i - 1][j + 1];
             int t = sum % 2;       /*已经是偶数了*/
             if(t == 0){
                 if(MAP[i][j] == 1)
                    return INF;
             }
             else{
                if(MAP[i][j] == 0){
                    MAP[i][j] = 1;
                    ans ++;
                }
             }
    }
    return ans;
}
int main(){
    int T;
    scanf("%d",&T);
    for(int Case = 1;Case <= T; Case ++){
        int ans = INF;
        scanf("%d",&n);
        for(int i = 0 ; i < n ; i++)
            for(int j = 0 ; j < n ; j++)
                scanf("%d",&Map[i][j]);
        for(int i = 0 ; i < (1 << n) ; i++)
            ans = min(ans,solve(i));
        printf("Case %d: ",Case);
        if(ans >= INF)
        printf("-1\n");
        else
        printf("%d\n",ans);
    }
    return 0;
}