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HDU 4101 Ali and Baba

2024-07-13 00:03:02   219人阅读

Ali and Baba

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1359    Accepted Submission(s): 287


Problem Description
There is a rectangle area (with N rows and M columns) in front of Ali and Baba, each grid might be one of the following:
1. Empty area, represented by an integer 0.
2. A Stone, represented by an integer x (x > 0) which denote the HP of this stone.
3. Treasure, represented by an integer -1.
Now, Ali and Baba get the map of this mysterious area, and play the following game:
Ali and Baba play alternately, with Ali starting. In each turn, the player will choose a stone that he can touch and hit it. After this operation, the HP of the stone that been hit will decrease by 1. If some stone’s HP is decreased to 0, it will become an empty area. Here, a player can touch a stone means
there is path consist of empty area from the outside to the stone. Note that two grids are adjacent if and only if they share an edge.
The player who hits the treasure first wins the game.
 

Input
The input consists several testcases.
The first line contains two integer N and M (0 < N,M <= 300), the size of the maze.
The following N lines each contains M integers (less than 100), describes the maze, where a positive integer represents the HP of a stone, 0 reperents an empty area, and -1 reperents the treasure.
There is only one grid contains the treasure in the maze.
 

Output
“Ali Win” or “Baba Win” indicates the winner of the game.
 

Sample Input
3 3
1 1 1
1 -1 1
1 1 1
 

Sample Output
Baba Win
 

Source
2011 Alibaba-Cup Campus Contest
 

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题意：给一个矩阵，0表示空的可走，-1宝藏的位置（仅仅有一个），其余的正整数表示该位置石头数。甲乙两个人，轮流玩游戏。甲先玩，轮到Ａ走的时候，宝藏和外面有一条路，则Ａ获胜。假设不能直接到达宝藏位置，能够从外面通过空到某个位置，拿走该位置上的一个石头，假设该位置没有石头则变成空的可走。


思路：找到包围宝藏的圈，然后统计圈外面有多少石头（圈上的石头变为１），假设为奇数，则Ali Win，否则Baba Win

先从宝藏ＢＦＳ找到圈边界，假设能够到达外面，则Ali Win，否则，从外面ＢＦＳ，假设是圈边界，则cnt+=n-1;否则假设是石头，则所有统计。然后推断奇偶；


遇到的问题：

刚開始思路错误，開始是直接统计圈里面多少石头，然后减去，然后推断奇偶。

ＮＭ范围看错了。看成了100

由于第二个BFS是复制的第一个BFS，导致里面有些错误没有注意到！！比方　应该是nx 结果是ｘ　导致ＷＡ了非常久


#include <stdio.h>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <set>

using namespace std;

//#define WIN
#ifdef WIN
typedef __int64 LL;
#define iform "%I64d"
#define oform "%I64d\n"
#define oform1 "%I64d"
#else
typedef long long LL;
#define iform "%lld"
#define oform "%lld\n"
#define oform1 "%lld"
#endif

#define S64I(a) scanf(iform, &(a))
#define P64I(a) printf(oform, (a))
#define S64I1(a) scanf(iform1, &(a))
#define P64I1(a) printf(oform1, (a))
#define FOR(i, s, t) for(int (i)=(s); (i)<(t); (i)++)

const int INF = 0x3f3f3f3f;
const double eps = 10e-9;
const double PI = (4.0*atan(1.0));

const int maxn = 300 + 20;
const int moveX[] = {-1, 1, 0, 0};
const int moveY[] = {0, 0, -1, 1};
int G[maxn][maxn];
int TG[maxn][maxn];
int vis[maxn][maxn];
int n, m;

struct Node {
    int x, y;
    Node(int xx=0, int yy=0) {
        x = xx;
        y = yy;
    }
};

queue<Node> Q;
int bfs(int sx, int sy) {
    memset(TG, 0, sizeof(TG));
    memset(vis, 0, sizeof(vis));
    while(!Q.empty()) Q.pop();
    Q.push(Node(sx, sy));
    vis[sx][sy] = 1;
    int res = 0;
    while(!Q.empty()) {
        int x = Q.front().x;
        int y = Q.front().y;
        Q.pop();
        for(int i=0; i<4; i++) {
            int nx = x + moveX[i];
            int ny = y + moveY[i];
            if(nx < 1 || nx > n || ny < 1 || ny > m) return 1;
            if(vis[nx][ny]) continue;
            vis[nx][ny] = 1;
            if(G[nx][ny] > 0) {TG[nx][ny] = 1;}
            else Q.push(Node(nx, ny));
        }
    }
    return 0;
}

int bfs1(int sx, int sy) {
    memset(vis, 0, sizeof(vis));
    while(!Q.empty()) Q.pop();
    Q.push(Node(sx, sy));
    vis[sx][sy] = 1;
    int res = 0;
    while(!Q.empty()) {
        int x = Q.front().x;
        int y = Q.front().y;
        Q.pop();
        for(int i=0; i<4; i++) {
            int nx = x + moveX[i];
            int ny = y + moveY[i];
            if(nx < 0 || nx > n+1 || ny < 0 || ny > m+1) continue;
            if(vis[nx][ny]) continue;
            vis[nx][ny] = 1;
            if(TG[nx][ny] == 1) {
                res += G[nx][ny] - 1;
                //printf("%d %d -> %d = %d\n", nx, ny, G[nx][ny]-1, res);
            } else {
                Q.push(Node(nx, ny));
                if(G[nx][ny] > 0) res += G[nx][ny];
                //if(G[nx][ny] > 0) printf("%d %d -> %d = %d\n", nx, ny, G[nx][ny], res);
            }   
        }
    }
    return res;
}

int main() {

    while(scanf("%d%d", &n, &m) != EOF) {
        if(!n || !m) while(1);
        /*if(!n || !m) {
            puts("Ali Win");
            continue;
        }*/
        int sum = 0;
        int sx, sy;
        memset(G, 0, sizeof(G));
        for(int i=1; i<=n; i++) {
            for(int j=1; j<=m; j++) {
                scanf("%d", &G[i][j]);
                if(G[i][j] > 0) sum += G[i][j];
                if(G[i][j] == -1) {
                    sx = i;
                    sy = j;
                }
            }
        }
        int t = bfs(sx, sy);
        if(t) {
            puts("Ali Win");
            continue;
        }
        t = bfs1(0, 0);
        //printf("-->%d\n", t);
        if(t&1) puts("Ali Win");
        else puts("Baba Win");
    }

    return 0;
}









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