UVaLive 6585 && Gym 100299F Draughts (暴力+回溯)

题意：给定一个 10*10的矩阵，每一个W可以跳过一个B向对角走到#并把B吃掉，并且可以一直跳直到不能动为止，现在是W走的时候，问你最多吃几个B。

析：直接暴力+回溯，深搜就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 26 + 5;
const int mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
char s[15][15];
int ans;

void dfs(int r, int c, int cnt){
    ans = Max(ans, cnt);
    for(int i = 4; i < 8; ++i){
        int x = r + dr[i];
        int y = c + dc[i];
        int xx = x + dr[i];
        int yy = y + dc[i];

        if(is_in(xx, yy) && s[x][y] == 'B' && s[xx][yy] == '#'){
            s[x][y] = '#';
            dfs(xx, yy, cnt+1);
            s[x][y] = 'B';
        }
    }
}

int main(){
    int T;  cin >> T;
    n = m = 10;
    while(T--){
        for(int i = 0; i < 10; ++i)  scanf("%s", s+i);

        ans = 0;
        for(int i = 0; i < 10; ++i)
            for(int j = 0; j < 10; ++j)
                if(s[i][j] == 'W'){ s[i][j] = '#';  dfs(i, j, 0); s[i][j] = 'W'; }

        printf("%d\n", ans);
    }
    return 0;
}