【LOJ6408】「ICPC World Finals 2018」随波逐流

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【题目链接】

• 点击打开链接

【思路要点】

• 枚举行宽， $O(N)$ 计算最长河流长度即可。
• 具体来说，可以作为一个空位的前驱的空位至多有 $2$ 个，且是单调的，因此使用 $TwoPointers$ 来进行一个简单 $dp$ 即可。
• 时间复杂度 $O(N^2|S|)$ 。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2505;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
char s[MAXN];
int n, a[MAXN], layer[MAXN], endp[MAXN], dp[MAXN];
int main() {
	read(n);
	int Llen = 0, Rlen = -1;
	for (int i = 1; i <= n; i++) {
		scanf("%s", s + 1);
		a[i] = strlen(s + 1);
		chkmax(Llen, a[i]);
		Rlen += a[i] + 1;
	}
	int ans = 0, Max = 0;
	for (int len = Llen; len <= Rlen; len++) {
		for (int i = 1; i <= n; i++) {
			if (i == 1 || endp[i - 1] + 1 + a[i] > len) layer[i] = layer[i - 1] + 1, endp[i] = a[i];
			else layer[i] = layer[i - 1], endp[i] = endp[i - 1] + a[i] + 1;
		}
		int res = 0, pos = 1;
		memset(dp, 0, sizeof(dp));
		for (int i = 1; i <= n; i++) {
			while (pos < i && (layer[pos] < layer[i] - 1 || layer[pos] == layer[i] - 1 && endp[pos] < endp[i] - 1)) pos++;
			if (layer[pos] == layer[i] - 1 && endp[pos] <= endp[i] + 1) chkmax(dp[i], dp[pos]);
			if (pos + 1 <= i && layer[pos + 1] == layer[i] - 1 && endp[pos + 1] <= endp[i] + 1) chkmax(dp[i], dp[pos + 1]);
			if (layer[i] != layer[i + 1]) dp[i] = 0;
			else dp[i]++;
			chkmax(res, dp[i]);
		}
		if (res > Max) ans = len, Max = res;
	}
	printf("%d %d\n", ans, Max);
	return 0;
}