HDU - 6223 Infinite Fraction Path——剪枝

版权声明：欢迎大家转载，转载请注明出处 https://blog.csdn.net/hao_zong_yin/article/details/82752809

对于两个位置i，j,他们的下一个位置分别为pi = (i*i+1)%n, pj = (j*j+1)%n, pi==pj的概率很高，因此只要把这种情况减掉就能过了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
const int maxn = 2e5 + 10;
struct Node {
    int dep;
    char c;
    int p;
    bool operator < (const Node &rhs) const {
        if (dep != rhs.dep) return dep > rhs.dep;
        if (c != rhs.c) return c < rhs.c;
        return p > rhs.p;
    }
};
char s[maxn];
char ans[maxn];
int main() {
    int t, ks = 0;
    scanf("%d", &t);
    while (t--) {
        int n;
        scanf("%d", &n);
        scanf("%s", s);
        priority_queue<Node> que;
        for (int i = 0; i < n; i++) que.push(Node{0, s[i], i});
        char maxv = 0;
        int predep = -1, prep = -1;
        while (!que.empty()) {
            Node node = que.top(); que.pop();
            if (node.dep >= n) break;
            if (node.dep != predep) {
                maxv = node.c;
            }
            else maxv = max(maxv, node.c);
            if (node.c < maxv) continue;
            if (node.dep == predep && node.p == prep) continue;
            predep = node.dep, prep = node.p;
            ans[node.dep] = node.c;
            int nex = (1LL*node.p*node.p+1)%n;
            que.push(Node{node.dep+1, s[nex], nex});
        }
        printf("Case #%d: ", ++ks);
        for (int i = 0; i < n; i++) printf("%c", ans[i]);
        puts("");
    }
    return 0;
}