11464 Even Parity

最新推荐文章于 2019-05-12 15:04:07 发布

原创最新推荐文章于 2019-05-12 15:04:07 发布 · 378 阅读

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本文介绍了一种解决迷宫游戏问题的算法实现，通过状态表示和动态规划思想，该算法能够在给定迷宫布局的情况下计算出使游戏进入目标状态所需的最小操作数。文章详细展示了算法流程、关键数据结构及其实现细节。

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#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cstdlib>
using namespace std;
const int INF = ~0U >> 1;
const int maxn = 20;
const int dx[] = {0, -1, 0, 1};
const int dy[] = {-1, 0, 1, 0};
int T, n, kase = 0;
int str[maxn][maxn], s[maxn][maxn];

void print() {
    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= n; ++j)
            printf("%d ", s[i][j]);
        printf("\n");
    }
}

int slove(int ss) {
    int ans = 0;
    for(int i = 1; i <= n; ++i) {
        if(ss & (1 << (i - 1))) {
            if(str[1][i] == 0) {
                s[1][i] = 1;
                ++ans;
            }
        } else {
            if(str[1][i] == 1) return INF;
        }
    }
    //printf("++++\n");
    bool ok = true;
    int nx, ny;
    for(int i = 1; i < n; ++i) {
        for(int j = 1; j <= n; ++j) {
            int cnt = 0;
            for(int k = 0; k < 3; ++k) {
                nx = dx[k] + i, ny = dy[k] + j;
                cnt += s[nx][ny];
            }
            nx = i + dx[3], ny = j + dy[3];
            if(cnt == 1 || cnt == 3) { /// jishu
                if(s[nx][ny] == 0) {
                    ++ans;
                    s[nx][ny] = 1;
                }
            } else {
                //if(ss == 0) printf("%d  %d****%d  %d****%d  %d\n", i, j, dx[3], dy[3], nx, ny);
                if(s[nx][ny] == 1) {
                    ok = false;
                    goto leap;
                }
            }
        }
    }
    //if(ss == 0) print();
leap:
    if(ok) return ans;
    else return INF;
}

int main() {
    scanf("%d", &T);
    while(T--) {
        memset(str, 0, sizeof(str));
        memset(s, 0, sizeof(s));
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= n; ++j)
                scanf("%d", &str[i][j]);
        memcpy(s, str, sizeof(str));
        //print();
        int ans = INF;
        for(int i = 0; i < (1 << n); ++i) {
            int d = slove(i);
            //printf("%d+++\n", d);
            if(ans > d) ans = d;
            memcpy(s, str, sizeof(str));
        }
        //print();
        //printf("%d----\n", ans);
        printf("Case %d: ", ++kase);
        if(ans == INF) printf("-1\n");
        else printf("%d\n", ans);
    }
    return 0;
}