HDU 4101 Ali and Baba

我不是女神ヾ 2022-03-27 14:26 66阅读 0赞

长着一张博弈脸的搜索题。给出n\*m的地图，有石头空地若干，还有一个宝藏。正整数表示石头的hp，0表示空地，-1表示宝藏，而且保证只有一个。AB轮流敲，每次从边界开始，选一个可达的石头，敲一下，石头的hp-1，hp为零就变成空地了。谁先敲到宝藏，谁赢。

如果从边界直接能走到宝藏，那么先手的A就赢了。否则呢，就会有一层石头保护着宝藏，谁先打破保护层，谁就输。所以保护层外的石头加上保护层边界上每个石头的hp-1就构成了他们能敲的石头总数。就是说把保护层边界上的各个点的hp都只剩下1。

5 7  
0 1 1 1 1 1 0  
1 0 0 0 0 0 1  
1 0 0 0 -1 0 1  
1 0 0 0 0 0 1  
0 1 1 1 1 1 0  
5 7  
0 1 1 1 1 1 0  
1 0 0 0 0 0 1  
1 0 1 0 -1 0 1  
1 0 0 0 0 0 1  
0 1 1 1 1 1 0  
5 7  
0 1 1 1 1 1 0  
1 0 0 0 0 0 1  
1 0 2 0 -1 0 1  
1 0 0 0 0 0 1  
0 1 1 1 1 1 0

这三组数据改变了我企图一遍bfs的想法。

#pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<vector>
    #include<string>
    #include<queue>
    #include<cmath>
    #include<map>
    ///LOOP
    #define REP(i, n) for(int i = 0; i < n; i++)
    #define FF(i, a, b) for(int i = a; i < b; i++)
    #define FFF(i, a, b) for(int i = a; i <= b; i++)
    #define FD(i, a, b) for(int i = a - 1; i >= b; i--)
    #define FDD(i, a, b) for(int i = a; i >= b; i--)
    ///INPUT
    #define RI(n) scanf("%d", &n)
    #define RII(n, m) scanf("%d%d", &n, &m)
    #define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
    #define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
    #define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
    #define RFI(n) scanf("%lf", &n)
    #define RFII(n, m) scanf("%lf%lf", &n, &m)
    #define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
    #define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)
    #define RS(s) scanf("%s", s)
    ///OUTPUT
    #define PN printf("\n")
    #define PI(n) printf("%d\n", n)
    #define PIS(n) printf("%d ", n)
    #define PS(s) printf("%s\n", s)
    #define PSS(s) printf("%s ", s)
    #define PC(n) printf("Case %d: ", n)
    ///OTHER
    #define PB(x) push_back(x)
    #define CLR(a, b) memset(a, b, sizeof(a))
    #define CPY(a, b) memcpy(a, b, sizeof(b))
    #define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}
    #define lson l, m, rt << 1
    #define rson m + 1, r, rt << 1 | 1
    
    using namespace std;
    typedef long long LL;
    typedef pair<int, int> P;
    const int MOD = 1e9+7;
    const int INFI = 1e9 * 2;
    const LL LINFI = 1e17;
    const double eps = 1e-6;
    const double pi = acos(-1.0);
    const int N = 333;
    const int M = 888;
    const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};
    
    struct point
    {
        int x, y;
        point(){};
        point(int a, int b){x = a, y = b;};
    }p;
    int a[N][N], vis[N][N];
    int n, m, ans;
    queue<point> q;
    
    void bfs2()
    {
        int tx, ty;
        while(!q.empty())q.pop();
        q.push(point(0, 0));
        while(!q.empty())
        {
            p = q.front();
            q.pop();
            REP(i, 4)
            {
                tx = p.x + move[i][0];
                ty = p.y + move[i][1];
                if(tx < 0 || ty < 0 || tx > n + 1 || ty > m + 1)continue;
                if(!vis[tx][ty])
                {
                    vis[tx][ty] = 2;
                    ans += a[tx][ty];
                    q.push(point(tx, ty));
                }
                else if(vis[tx][ty] == 1)
                {
                    vis[tx][ty] = 2;
                    ans += (a[tx][ty] - 1);
                }
            }
        }
    }
    
    bool bfs()
    {
        int tx, ty;
        while(!q.empty())
        {
            p = q.front();
            q.pop();
            REP(i, 4)
            {
                tx = p.x + move[i][0];
                ty = p.y + move[i][1];
                if(tx <= 0 || ty <= 0 || tx > n || ty > m)return 0;
                if(!vis[tx][ty])
                {
                    vis[tx][ty] = 1;
                    if(!a[tx][ty])q.push(point(tx, ty));
                }
            }
        }
        bfs2();
        return 1;
    }
    
    int main()
    {
        //freopen("input.txt", "r", stdin);
    
        int x, y;
        while(RII(n, m) != EOF)
        {
            ans = 0;
            CLR(a, 0);
            CLR(vis, 0);
            FFF(i, 1, n)FFF(j, 1, m)
            {
                RI(a[i][j]);
                if(a[i][j] == -1)x = i, y = j;
            }
            while(!q.empty())q.pop();
            q.push(point(x, y));
            vis[x][y] = 1;
            if(bfs())puts((ans & 1) ? "Ali Win" : "Baba Win");
            else puts("Ali Win");
        }
        return 0;
    }