哈尔滨理工大学软件与微电子学院第八届程序设计竞赛同步赛（高年级）

https://ac.nowcoder.com/acm/contest/301#question

 

F.小乐乐下象棋

 

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

const long long mod = 1000000007;
int n, m, k;
int sx, sy;
int ex, ey;
long long dp[220][220][220];

int dir[8][2] = {
    {1, 2},
    {2, 1},
    {2, -1},
    {1, -2},
    {-1, -2},
    {-2, -1},
    {-2, 1},
    {-1, 2}
};

int out(int x, int y) {
    if(x <= 0 || x > n) return 1;
    if(y <= 0 || y > m) return 1;
    return 0;
}

int main() {
    while(~scanf("%d%d%d", &n, &m, &k)) {
        memset(dp, 0, sizeof(dp));

        dp[1][1][0] = 1;
        for(int step = 1; step <= k; step ++) {
            for(int i = 1; i <= n; i ++) {
                for(int j = 1; j <= m; j ++) {
                    for(int d = 0; d < 8; d ++) {
                        int ii = i + dir[d][0];
                        int jj = j + dir[d][1];
                        if(out(ii, jj)) continue;
                        dp[i][j][step] = (dp[i][j][step] + dp[ii][jj][step - 1]) % mod;
                    }
                }
            }
        }

        cout << dp[n][m][k] << endl;
    }
    
    return 0;
}

（判断是否可以 k 步走到还可以剪枝 明天再写）

#include <bits/stdc++.h>
using namespace std;

int n, m, k;
int mp[110][110];
int vis[110][110];
int sx, sy;
int ex, ey;
bool flag = false;

int dir[8][2] = {
    {1, 2},
    {2, 1},
    {2, -1},
    {1, -2},
    {-1, -2},
    {-2, -1},
    {-2, 1},
    {-1, 2}
};

void dfs(int x, int y, int step) {
    if(x == ex && y == ey && step == k) {
        flag = true;
        return ;
    }

    for(int i = 0; i < 8; i ++) {
        int xx = x + dir[i][0];
        int yy = y + dir[i][1];

        if(xx >= 0 && xx < n && yy >= 0 && yy < m && step < k) {
            vis[xx][yy] = 1;
            dfs(xx, yy, step + 1);
            vis[xx][yy] = 0;
        }
    }
}

int main() {
    scanf("%d%d%d", &n, &m, &k);
    sx = 0, sy = 0;
    ex = n - 1, ey = m - 1;

    memset(vis, 0, sizeof(vis));
    vis[sx][sy] = 1;

    dfs(0, 0, 0);
    if(flag) printf("yes\n");
    else printf("no\n");
    return 0;
}