Luogu P4907 A换B problem

题目传送门

暴力乱搞大法好！

感觉这题难度有点虚，最多蓝题吧

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一开始写了个没有一丝剪枝的搜索交了上去，水到了\(60pts\)
于是加了个卡时，就\(A\)了

顺便说一句，数据略水（可能也是因为乱搞不好卡吧）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <ctime>
#define LL long long
using namespace std;
LL read() {
    LL k = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9')
        k = k * 10 + c - 48, c = getchar();
    return k * f;
}
bool tong[5][100]; //tong[花色][点数]
struct zzz {
    int c, k;
}a[100]; //c:花色，k:点数
bool flag;
int n, ans = 100, ans2 = 100, kti;
bool judge() { //判断是否为顺子
    int num = 0; bool flag2 = 1;
    for(int i = 1; i <= 4; ++i) {
        int minn = 100, maxn = 0;
        for(int j = 1; j <= 13; ++j) {
            if(tong[i][j] && minn == 100) minn = j;
            if(tong[i][j]) maxn = max(maxn, j);
        }
        for(int j = minn; j <= maxn; ++j)
            if(!tong[i][j]) flag2 = 0, ++num;
    }
    ans2 = min(ans2, num);
    return flag2;
}
void dfs(int pos, int tot) {
    if(tot >= ans) return ;
    if(judge()) {
        ans = min(ans, tot);
        return ;
    }
    if(pos > n) return ;
    //=====卡时
    if((clock() - kti) * 1000 >= 980 * CLOCKS_PER_SEC){
        if(ans <= 52) printf("Yes\n%d", ans);
        else printf("No\n%d", ans2);
        exit(0);
    }
    //=====
    for(int i = 1; i <= 4; ++i) {
        if(tong[i][a[pos].k] && i != a[pos].c) continue;
        tong[a[pos].c][a[pos].k] = 0;
        tong[i][a[pos].k] = 1;
        if(i != a[pos].c) dfs(pos+1, tot+1);
        else dfs(pos+1, tot);
        tong[i][a[pos].k] = 0, tong[a[pos].c][a[pos].k] = 1;
    }
}
int main() {
    kti = clock(), n = read();
    for(int i = 1; i <= n; ++i) {
        a[i].c = read();
        char c[2]; cin >> c;
        if(c[0] == 'A') a[i].k = 1;
        else if(c[0] == 'J') a[i].k = 11;
        else if(c[0] == 'Q') a[i].k = 12;
        else if(c[0] == 'K') a[i].k = 13;
        else if(c[0] == '1') a[i].k = 10;
        else a[i].k = c[0] - '0';
    }
    for(int i = 1; i <= n; ++i) tong[a[i].c][a[i].k] = 1;
    dfs(1, 0);
    if(ans <= 52) printf("Yes\n%d", ans);
    else printf("No\n%d", ans2);
    return 0;
}