POJ 1753 Flip Game 高斯消元

先用高斯消元化简,然后枚举自由变元

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <climits>
#include <iostream>
#include <string>

using namespace std;
 
#define MP make_pair
#define PB push_back
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int INF = INT_MAX / 3;
const double eps = 1e-8;
const LL LINF = 1e17;
const double DINF = 1e60;
const int maxn = 250;
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
int a[maxn][maxn], n;
bool fx[maxn];
char buf[16][16];

void Gauss() {
    for(int i = 0; i < n * n; i++) {
        int k = i;
        while(a[k][i] == 0 && k < n * n) k++;
        if(k >= n * n) break;
        for(int j = 0; j <= n * n; j++) swap(a[i][j], a[k][j]);
        for(int j = 0; j < n * n; j++) if(j != i && a[j][i] != 0) {
            for(int k = 0; k <= n * n; k++) {
                a[j][k] ^= a[i][k];
            }
        }
    }
}

void pp() {
    for(int i = 0; i < n * n; i++) {
        for(int j = 0; j <= n * n; j++) {
            printf("%d ", a[i][j]);
        }
        puts("");
    }
}

int solve() {
    memset(a, 0, sizeof(a));
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            if(buf[i][j] == 'w') a[i * n + j][n * n] = 1;
            else a[i * n + j][n * n] = 0;
        }
    }
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            int u = i * n + j;
            a[u][u] = 1;
            for(int k = 0; k < 4; k++) {
                int nx = i + dx[k], ny = j + dy[k], nu = nx * n + ny;
                if(nx >= 0 && nx < n && ny >= 0 && ny < n) {
                    a[u][nu] = a[nu][u] = 1;
                }
            }
        }
    }
    Gauss();
    memset(fx, 0, sizeof(fx));
    int ans = INF, fcnt = 0;
    bool bad = false;
    for(int i = 0; i < n * n; i++) {
        int nsum = 0;
        for(int j = 0; j  < n * n; j++) nsum += a[i][j];
        if(nsum == 0 && a[i][n * n] == 1) bad = true;
        if(nsum == 0) fx[i] = true, fcnt++;
    }
    if(bad) return INF;
    for(int s = 0; s < (1 << fcnt); s++) {
        int nowans = 0;
        for(int i = 0; i < n * n - fcnt; i++) {
            int col = 0;
            for(int j = 0; j < n * n; j++) if(fx[j] && a[i][j]) {
                if(s & (1 << (n * n - j - 1))) col ^= 1;
            }
            if(a[i][n * n] != col) nowans++; 
        }
        int bcnt = 0;
        for(int i = 0; i < fcnt; i++) if(s & (1 << i)) bcnt++;
        ans = min(nowans + bcnt, ans);
    }
    return ans;
}

int main() {
    while(scanf("%s%s%s%s", buf[0], buf[1], buf[2], buf[3]) != EOF) {
        n = 4;
        int ans = solve();
        for(int i = 0; i < 4; i++) {
            for(int j = 0; j < 4; j++) {
                if(buf[i][j] == 'w') buf[i][j] = 'b';
                else buf[i][j] = 'w';
            }
        }
        ans = min(ans, solve());
        if(ans < INF) printf("%d
", ans);
        else puts("Impossible");
    }
    return 0;
}

  

原文地址:https://www.cnblogs.com/rolight/p/4066466.html