CCF CSP——202206-2 寻宝！大冒险！

问题描述：

100分题解

题解描述：此题即为在一个大图里面匹配一个小图问题，枚举出每一种可能的情况即可；优化方式，因树的坐标相对较少，并且题目给的输入也是树的坐标，因此只需要保存树的坐标，将大图（绿化图）的树的坐标与小图（藏宝图）的树的坐标对应之后，枚举每个点匹配即可。

值得注意的一点是遍历藏宝图的坐标时使用一层for循环代替两层for循环为了当匹配失败时一个break即可提前终止循环

python题解

n, L, s = map(int, input().split())
points = set()
# 保存绿化图树的坐标
for i in range(n):
    p = tuple(map(int, input().split()))
    points.add(p)
# 保存藏宝图树的坐标
money = set()
for i in range(s, -1, -1):
    ps = input().split()
    for j in range(s + 1):
        if ps[j] == '1':
            p = i, j
            money.add(p)
# 枚举匹配
ans = 0
for x, y in points:
    for i in range((s + 1) ** 2):
        temp = i // (s + 1), i % (s + 1)
        p = temp[0] + x, temp[1] + y
        if temp in money:
            if p not in points:
                break
        elif p in points or p[0] > L or p[1] > L:
            break
    else:
        ans += 1
print(ans)

c++题解

#include <iostream>
#include <vector>
using namespace std;

int main() {
    int n,L,s;
    cin >>n>>L>>s;
    int x,y,temp;
    int points[n][2];
    for (int i = 0; i < n; ++i) {
        cin >> points[i][0] >> points[i][1];
    }

    vector<int> money[2];
    for (int i = s; i >= 0; --i) {
        for (int j = 0; j <= s; ++j) {
            cin >> temp;
            if(temp == 1){
                money[0].push_back(i);
                money[1].push_back(j);
            }
        }
    }

    int ans = 0,sx,sy;
    for (int i = 0; i < n; ++i) {
        sx = points[i][0];
        sy = points[i][1];
        int k;
        for (k = 1; k < (s+1) * (s+1); ++k) {
            x = k / (s+1) + sx;
            y = k % (s+1) + sy;
            if(x > L || y > L)
                break;
            bool flag1 = false,flag2 = false;
            for (int m = 0; m < n; ++m)
                if(x == points[m][0] && y == points[m][1]){
                    flag1 = true;
                    break;
                }
            for (int m = 0; m < money[1].size(); ++m) {
                if(k / (s+1) == money[0][m] && k % (s+1) == money[1][m]){
                    flag2 = true;
                    break;
                }
            }
            if(flag1 ^ flag2)
                break;
        }
        if(k == (s+1) * (s+1))
            ans++;
    }
    cout << ans << endl;
    return 0;
}