洛谷 P9868 [NOIP2023] 词典

好久不写博客了，今天来水一篇

原题链接

初看此题在洛谷上的定位是黄题，实际上也并不是很简单。

其实主要就用到了贪心的思想，先说一下我在做题的时候是怎么想的吧。

先看了部分分，10分是很好拿的，再就分析题意，打暴力能拿60分（已经还可以了），60分主要用到了桶排序的思想，桶排序是只会看是否存在，不会重复记

60pts：（缺少了判断i！=j的情况）

#include<bits/stdc++.h>
using namespace std;
int read() {
	int x = 0, f = 1;
	char ch = getchar();
	while (ch < '0' || ch > '9') {
		if (ch == '-') f = -1;
		ch = getchar();
	}
	while (ch <= '9' && ch >= '0') {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	return x * f;
}
const int N = 3010;
string s[N];
int cnt[26];
int n, m;
int main() {
	n = read(), m = read();
	string mins = "";
	for (int i = 0; i < 3000; i++) {
		mins += "z";
	}
	for (int i = 1; i <= n; i++) {
		cin >> s[i];
		if (mins > s[i]) {
			mins = s[i];
		}
	}
	for (int i = 1; i <= n; i++) {
		int len = s[i].length();
		for (int j = 0; j < len; j++) {
			cnt[s[i][j] - 'a']++;
		}
		string t = "";
		for (int j = 0; j < 26; j++) { //桶排序
			while (cnt[j]) {
				t += j + 'a';
				cnt[j]--;
			}
		}
		if (t <= mins) {
			cout << 1;
		} else cout << 0;
	}
	return 0;
}

接下来就是正解：直接把每个字符串wi都从小到大或者从大到小排一下，记作ai，bi。如果bi小于除了i之外的所有ai，说明可以，否则不可以。求一个前后缀最大值即可。复杂度为O（26n+nm）

100pts：

#include<bits/stdc++.h>
using namespace std;
inline int read() {
	int x = 0, f = 1;
	char ch = getchar();
	while (ch < '0' || ch > '9') {
		if (ch == '-')f = -1;
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9')x = x * 10 + ch - '0', ch = getchar();
	return x * f;
}
const int N=3005;
char s[N];
char mx[N][N], mn[N][N];
char  pr[N][N], sf[N][N];
int c[35];
int  main() {
	int n = read(), m = read();
	for (int i = 1; i <= n; i++) {
		scanf("%s", s);
		int pmx = 0, pmn = 0;
		for (int j = 0; j < m; j++)c[s[j] - 'a']++;
		for (int j = 25; j >= 0; j--) {
			while (c[j])mx[i][pmx++] = j + 'a', c[j]--;
		}
		for (int j = 0; j < m; j++)c[s[j] - 'a']++;
		for (int j = 0; j <= 25; j++) {
			while (c[j])mn[i][pmn++] = j + 'a', c[j]--;
		}
	}
	for (int j = 0; j < m; j++)pr[1][j] = mx[1][j];
	for (int i = 2; i <= n; i++) {
		int flag = 0;
		for (int j = 0; j < m; j++) {
			if (pr[i - 1][j] < mx[i][j]) {
				flag = 0;
				break;
			} else if (pr[i - 1][j] > mx[i][j]) {
				flag = 1;
				break;
			}
		}
		if (!flag) {
			for (int j = 0; j < m; j++)pr[i][j] = pr[i - 1][j];
		} else {
			for (int j = 0; j < m; j++)pr[i][j] = mx[i][j];
		}
	}
	for (int j = 0; j < m; j++)sf[n][j] = mx[n][j];
	for (int i = n - 1; i >= 1; i--) {
		int flag = 0;
		for (int j = 0; j < m; j++) {
			if (sf[i + 1][j] < mx[i][j]) {
				flag = 0;
				break;
			} else if (sf[i + 1][j] > mx[i][j]) {
				flag = 1;
				break;
			}
		}
		if (!flag) {
			for (int j = 0; j < m; j++)sf[i][j] = sf[i + 1][j];
		} else {
			for (int j = 0; j < m; j++)sf[i][j] = mx[i][j];
		}
	}
	for (int i = 1; i <= n; i++) {
		int flag = 1;
		if (i > 1) {
			int tag = 0;
			for (int j = 0; j < m; j++) {
				if (mn[i][j] > pr[i - 1][j]) {
					tag = 0;
					break;
				} else if (mn[i][j] < pr[i - 1][j]) {
					tag = 1;
					break;
				}
			}
			flag &= tag;
		}
		if (i < n) {
			int tag = 0;
			for (int j = 0; j < m; j++) {
				if (mn[i][j] > sf[i + 1][j]) {
					tag = 0;
					break;
				} else if (mn[i][j] < sf[i + 1][j]) {
					tag = 1;
					break;
				}
			}
			flag &= tag;
		}
		if (flag)cout<<1;
		else cout<<0;
	}
	return 0;
}