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【考前模拟】正确答案

1.正确答案

【题目描述】

小H与小Y刚刚参加完UOIP外卡组的初赛,就迫不及待的跑出考场对答案。

“吔,我的答案和你都不一样!”,小Y说道,”我们去找神犇们问答案吧”。

外卡组试卷中共有m道判断题,小H与小Y一共从其他n个神犇那问了答案。之后又从小G那里得知,这n个神犇中有p个考了满分,q个考了零分,其他神犇不为满分或零分。这可让小Y与小H犯了难。你能帮助他们还原出标准答案吗?如有多解则输出字典序最小的那个。无解输出-1。

 

【输入格式】

第一行四个整数n, m, p, q,意义如上描述。

接下来n行,每一行m个字符’N’或’Y’,表示这题这个神犇的答案。

 

【输出格式】

仅一行,一个长度为m的字符串或是-1。

 

【样例输入】

2 2 2 0

YY

YY

 

【样例输出】

YY

 

【数据范围】

30% : n <= 100.

60% : n <= 5000 , m <= 100.

100% : 1 <= n <= 30000 , 1 <= m <= 500.  0 <= p , q 且 p + q <= n.

 

T1:

30%:  O(n ^ 2 * m)暴力判断。

100%: 很显然答案的可能性最多只有n种,所以我们将所有人的答案按字典序排序后枚举     将每个人的答案作为正确答案来进行判断。由于是判断题,若当前人的答案为正确答      案则零分者的答案也就确定了,那么只需统计出这两种答案的人数判断是否满足题意      即可。这一步使用字符串哈希即可解决。

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

const int N = 3e4 + 2, M = 5e2 + 2, sed = 31, SED = 131, mod = 70177, MOD = 92311;
int n, m, p, q, ans, hash[N], HASH[N];
int top, info[mod], nxt[N * 2], fet[N * 2], cnt[N * 2];
struct node {
    char s[M];
    inline bool operator < (const node &b) const {
        return strcmp(s, b.s) < 0;
    }
} a[N];

inline void Insert(const int &x, const int &y) {
    for (int k = info[x]; k; k = nxt[k])
        if (fet[k] == y) {
            ++cnt[k]; return ;
        }
    nxt[++top] = info[x]; info[x] = top;
    fet[top] = y; cnt[top] = 1;
    return ;
}

inline int Query(const int &x, const int &y) {
    for (int k = info[x]; k; k = nxt[k])
        if (fet[k] == y) return cnt[k];
    return 0;
}

inline void Solve1() {
    int tmp, TMP; ans = -1;
    for (int i = 0; i < n; ++i) {
        tmp = TMP = 0;
        for (int j = 0; j < m; ++j) {
            tmp = (tmp * sed + (a[i].s[j] == N)) % mod;
            TMP = (TMP * SED + (a[i].s[j] == N)) % MOD;
        }
        hash[i] = tmp, HASH[i] = TMP;
        Insert(tmp, TMP);
    }
    for (int i = 0; i < n; ++i)
        if (Query(hash[i], HASH[i]) == p) {
            tmp = TMP = 0;
            for (int j = 0; j < m; ++j) {
                tmp = (tmp * sed + (a[i].s[j] == Y)) % mod;
                TMP = (TMP * SED + (a[i].s[j] == Y)) % MOD;
            }
            if (Query(tmp, TMP) == q) {
                ans = i; break;
            }
        }
    if (ans != -1) printf("%s\n", a[ans].s);
    else     puts("-1");
    return ;
}

char cur[M];
inline void Solve2() {
    int tmp, TMP; ans = -1;
    for (int i = 0; i < n; ++i) {
        tmp = TMP = 0;
        for (int j = 0; j < m; ++j) {
            tmp = (tmp * sed + (a[i].s[j] == N)) % mod;
            TMP = (TMP * SED + (a[i].s[j] == N)) % MOD;
        }
        hash[i] = tmp, HASH[i] = TMP;
        Insert(tmp, TMP);
    }
    for (int i = n - 1; i >= 0; --i)
        if (Query(hash[i], HASH[i]) == q) {
            tmp = TMP = 0;
            for (int j = 0; j < m; ++j) {
                tmp = (tmp * sed + (a[i].s[j] == Y)) % mod;
                TMP = (TMP * SED + (a[i].s[j] == Y)) % MOD;
            }
            if (Query(tmp, TMP) == p) {
                ans = i; break;
            }
        }
    if (ans != -1) {
        for (int i = 0; i < m; ++i)
            cur[i] = a[ans].s[i] == N ? Y : N;
        printf("%s\n", cur);
    }
    else     puts("-1");
    return ;
}

void Solve3() {
    int tmp, TMP;
    for (int i = 0; i < n; ++i) {
        tmp = TMP = 0;
        for (int j = 0; j < m; ++j) {
            tmp = (tmp * sed + (a[i].s[j] == N)) % mod;
            TMP = (TMP * SED + (a[i].s[j] == N)) % MOD;
        }
        Insert(tmp, TMP);
        tmp = TMP = 0;
        for (int j = 0; j < m; ++j) {
            tmp = (tmp * sed + (a[i].s[j] == Y)) % mod;
            TMP = (TMP * SED + (a[i].s[j] == Y)) % MOD;
        }
        Insert(tmp, TMP);
    }
    bool flag = true;
    for (int i = 0; i < m; ++i) cur[i] = N;
    do {
        tmp = TMP = 0;
        for (int j = 0; j < m; ++j) {
            tmp = (tmp * sed + (cur[j] == N)) % mod;
            TMP = (TMP * SED + (cur[j] == N)) % MOD;
        }
        if (Query(tmp, TMP) == 0) {
            flag = true; break;
        }
        flag = false;
        for (int j = m - 1; j >= 0; --j)
            if (cur[j] == Y) cur[j] = N;
            else {
                cur[j] = Y; flag = true; break;
            }
    } while (flag);
    if (flag) printf("%s\n", cur);
    else     puts("-1");
    return ;
}

int main() {
    freopen("answer.in", "r", stdin);
    freopen("answer.out", "w", stdout);
    scanf("%d%d%d%d", &n, &m, &p, &q);
    for (int i = 0; i < n; ++i) scanf("%s", a[i].s);
    sort(a, a + n);
    if (p) Solve1();
    else if (q) Solve2();
    else     Solve3();
    fclose(stdin); fclose(stdout);
    return 0;
}

 

【考前模拟】正确答案