vijos 1218 数字游戏(环形dp)

第一次接触环形的dp，学到了一招:把环复制一次就变成了线了,不过复杂度同样也要上一层.

枚举每一个切割点k,对[k, k + n]序列做如下dp:

设dp[i][j]为前i个数字分成j堆能得到最大/最小值.

则dp[i][j] = max/min{dp[p][j - 1] * (sum(p + 1, i) % 10) | 1 <= p < i}

base case:dp[i][1] = sum(1, i)

这里的sum(x, y)指从x开始到y的序列和

复杂度为mn^3

#include <cstdio>
#include <memory.h>
#include <climits>
#include <algorithm>
using namespace std;
const int MAX = 102;
int dp_min[51][10], dp_max[51][10];
int num[MAX], s[MAX];

int sum(int x, int y){
	return s[y] - s[x - 1];
}
int mod(int a, int m){
	return (a % m + m) % m;
}
int main(int argc, char const *argv[]){
	int n, m, min_ans = INT_MAX, max_ans = INT_MIN;
	
	scanf("%d%d", &n, &m);
	for(int i = 0; i < n; ++i){
		scanf("%d", &num[i]);
		num[i + n] = num[i];
	}
	
	s[0] = num[0];
	for(int i = 1; i < n * 2; ++i){
		s[i] = s[i - 1] + num[i];
	}

	//turn the circle into a line by cutting at kth number
	for(int k = 0; k < n; ++k){
		for(int i = 0; i < n; ++i){
			dp_min[i][1] = mod(sum(k, i + k), 10);
			dp_max[i][1] = mod(sum(k, i + k), 10);
		}
		for(int i = 0; i < n; ++i){
			for(int j = 2; j <= m && j <= i + 1; ++j){
				int tmin = INT_MAX, tmax = INT_MIN;
				for(int p = 0; p < i; ++p){
					int v1 = dp_min[p][j - 1] * mod(sum(k + p + 1, k + i), 10);
					int v2 = dp_max[p][j - 1] * mod(sum(k + p + 1, k + i), 10);
					tmin = min(tmin, v1);
					tmax = max(tmax, v2);
				}
				dp_min[i][j] = tmin;
				dp_max[i][j] = tmax;
			}
		}
		min_ans = min(min_ans, dp_min[n - 1][m]);
		max_ans = max(max_ans, dp_max[n - 1][m]);
	}
	printf("%d\n%d\n", min_ans, max_ans);
	return 0;
}

vijos 1218 数字游戏(环形dp),布布扣,bubuko.com

vijos 1218 数字游戏(环形dp)

原文：http://blog.csdn.net/zxjcarrot/article/details/21820653