标签：acm   c++   dp   

Big Event in HDU

2
10 1
20 1
3
10 1 
20 2
30 1
-1

20 10
40 40

代码：

/**
*   本题题意：
*           给出每种石头的重量和个数，求尽量将这些石头分成两堆时，两堆石块的重量
*           这里消耗和价值都是石块的重量
*   解题思路：V=sum/2
*/
#include <string.h>
#include <stdio.h>
#define maxn 500005
int max(int a, int b){
	if (a > b)return a;
	return b;
}
int dp[maxn];
int weight[maxn];
int number[maxn];
int V;
//01
void zeroonepack(int c)
{
	for (int v = V; v >= c; v--)
		dp[v] = max(dp[v], dp[v - c] + c);
}
//完全
void completepack(int c)
{
	for (int v = c; v <= V; v++)
		dp[v] = max(dp[v], dp[v - c] + c);
}
//多重
void mupack(int c, int p)
{
	if (c*p >= V)
		completepack(c);
	else
	{
		int k = 1;
		while (k < p)
		{
			zeroonepack(k*c);
			p = p - k;
			k = k * 2;
		}
		zeroonepack(p*c);
	}
}
int main()
{
	int n;
	while (scanf("%d", &n), n >= 0)
	{
		int sum = 0;
		for (int i = 1; i <= n; i++)
		{
			scanf("%d%d", &weight[i], &number[i]);
			sum += weight[i] * number[i];
		}
		V = sum / 2;
		memset(dp, 0, sizeof(dp));
		for (int i = 1; i <= n; i++)
			mupack(weight[i], number[i]);
		int x = dp[V];
		int y = sum - x;
		if (x > y)
			printf("%d %d\n", x, y);
		else
			printf("%d %d\n", y, x);
	}
	return 0;
}

hdu 1171 Big Event in HDU 多重背包,布布扣,bubuko.com

hdu 1171 Big Event in HDU 多重背包

标签：acm   c++   dp   

原文地址：http://blog.csdn.net/u012964281/article/details/26400161