【BZOJ1044】[HAOI2008]木棍分割

【BZOJ1044】[HAOI2008]木棍分割

题面

bzoj

洛谷

题解

第一问显然可以二分出来的。

第二问：

设\(dp[i][j]\)表示前\(i\)个，切了\(j\)组的方案数

发现每次转移都是从前面一个区间过来的

直接前缀和优化就好了

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring> 
#include <cmath> 
#include <algorithm>
using namespace std; 
inline int gi() {
	register int data = 0, w = 1; 
	register char ch = 0; 
	while (!isdigit(ch) && ch != '-') ch = getchar(); 
	if (ch == '-') w = -1, ch = getchar(); 
	while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar(); 
	return w * data; 
}
const int Mod = 10007;
void pls(int &x, int y) { x += y; if (x >= Mod) x -= Mod; }
void dec(int &x, int y) { x -= y; if (x < 0) x += Mod; } 
const int MAX_N = 50005; 
int N, M, ans1, ans2, len[MAX_N], L[MAX_N], prv[MAX_N]; 
bool check(int v) {
	int cnt = 0, s = 0; 
	for (int i = 1; i <= N; i++) { 
		if (s + len[i] > v) ++cnt, s = len[i]; 
		else s += len[i]; 
		if (cnt > M) return 0; 
	} 
	return 1; 
} 
int solve() { 
	int l = *max_element(&len[1], &len[N + 1]), r = L[N], res = L[N]; 
	while (l <= r) {
		int mid = (l + r) >> 1; 
		if (check(mid)) r = mid - 1, res = mid;
		else l = mid + 1; 
	}
	return res; 
} 
int sum1[MAX_N], sum2[MAX_N]; 
int main () {
	N = gi(), M = gi();
	for (int i = 1; i <= N; i++) L[i] = L[i - 1] + (len[i] = gi()); 
	ans1 = solve(); 
	for (int i = 1; i <= N; i++) { 
		int l = 1, r = i; prv[i] = i; 
		while (l <= r) {
			int mid = (l + r) >> 1; 
			if (L[i] - L[mid - 2] <= ans1) prv[i] = mid, r = mid - 1; 
			else l = mid + 1; 
		} 
	}
	for (int i = 1, j = 1; i <= N; i++) {
		while (L[i] - L[j - 1] > ans1 && j < i) ++j;
		if (L[i] - L[j - 1] <= ans1) prv[i] = max(j - 2, 0); 
	} 
	for (int i = 1; i <= N; i++) sum1[i] = sum1[i - 1], pls(sum1[i], (L[i] <= ans1)); 
	for (int i = 1; i <= M; i++) { 
		for (int j = 1; j <= N; j++) sum2[j] = sum1[j - 1], dec(sum2[j], sum1[prv[j]]); 
		for (int j = 1; j <= N; j++) sum1[j] = sum1[j - 1], pls(sum1[j], sum2[j]); 
		pls(ans2, sum2[N]); 
	}
	printf("%d %d\n", ans1, ans2); 
	return 0; 
}