POJ1837 Balance 背包

题目大意: 有一个天平,天平左右两边各有若干个钩子,总共有C个钩子(每个钩子有相对于中心的距离,左负右正),有G个钩码,求将钩码全部挂到钩子上使天平平衡的方法的总数。

将每个砝码看作一组,组内各个物品的体积为每个挂钩与该砝码形成的力矩,背包总体积严格为0,这便是分组背包计数问题(特殊点:每一组必须出一个物品,而不是至多出一个物品)。由于c++不允许负的数组下标,所以每次更新时,j要加上offsetJ。

实现分组背包计数问题时,可以用填表法(找以前节点求自己值)(DP1)或刷表法(找以后节点更新以后值)(DP2)。由于刷表法时,如果DP[i][j]==0,可以跳过,所以节省时间。

注意:

  • 不可以用一维数组倒序循环来表示DP数组,因为j-objV可能比j还大
  • 同理,每次判断时,不是j-objV>=0,0代表天平的支点而不是左端点。所以应当为j-objV>=minJ。
#include <cstdio>
#include <cstring>
#include <cstdarg>
using namespace std;

const int MAX_V = 16000, MAX_OBJ = 30, MAX_HOOP = 30;
const int Plus = 7500;
#define Sub(x) x+offsetJ
int TotHoop;
int Len[MAX_HOOP], W[MAX_OBJ];

void _printf(char *format, ...)
{
#ifdef _DEBUG
    va_list(args);
    va_start(args, format);
    vprintf(format, args);
    va_end(args);
#endif
}

int DP1(int totHoop, int totDev, int *_w, int *_len)
{
    int minJ = 0, maxJ = 0, wSum = 0, offsetJ, objV = 0;
    static int DP[MAX_OBJ][MAX_V];
    for (int dev = 1; dev <= totDev; dev++)
        wSum += _w[dev];
    for (int hoop = 1; hoop <= totHoop; hoop++)
        _len[hoop] > 0 ? maxJ += _len[hoop] * wSum : minJ += _len[hoop] * wSum;
    offsetJ = -minJ;
    DP[0][offsetJ] = 1;
    _printf("+ offsetJ %d maxJ %d
", +offsetJ, maxJ);
    for (int i = 1; i <= totDev; i++)
        for (int j = minJ; j <= maxJ; j++)
            for (int hoop = 1; hoop <= totHoop; hoop++)
            {
                objV = _w[i] * _len[hoop];
                if (j - objV >= minJ && j - objV <= maxJ)
                {
                    DP[i][j + offsetJ] += DP[(i - 1)][j + offsetJ - objV];
                    _printf("%d=DP[%d][%d] += DP[%d][%d]=%d
", DP[i][j + offsetJ], i, j, i - 1, j - _w[i] * _len[hoop], DP[i - 1][j - _w[i] * _len[hoop] + offsetJ]);
                }
            }
    return DP[totDev][offsetJ];
}

int DP2(int totHoop, int totDev, int *_w, int *_len)
{
    int minJ = 0, maxJ = 0, wSum = 0, offsetJ, objV = 0;
    static int DP[MAX_OBJ][MAX_V];
    for (int dev = 1; dev <= totDev; dev++)
        wSum += _w[dev];
    for (int hoop = 1; hoop <= totHoop; hoop++)
        _len[hoop] > 0 ? maxJ += _len[hoop] * wSum : minJ += _len[hoop] * wSum;
    offsetJ = -minJ;
    DP[0][offsetJ] = 1;
    _printf("+ offsetJ %d maxJ %d
", + offsetJ, maxJ);
    for (int i = 0; i < totDev; i++)
        for (int j = minJ; j <= maxJ; j++)
            if (DP[i][j+ offsetJ])
                for (int hoop = 1; hoop <= totHoop; hoop++)
                {
                    objV = _w[i+1] * _len[hoop];
                    DP[i + 1][j + objV + offsetJ] += DP[i][j + offsetJ];
                    _printf("%d=DP[%d][%d] += DP[%d][%d]=%d
", DP[i+1][j+objV+ offsetJ], i+1, j+objV+ offsetJ, i, j+ offsetJ , DP[i][j + offsetJ]);
                }
    return DP[totDev][offsetJ];
}

int main()
{
#ifdef _DEBUG
    freopen("c:\noi\source\input.txt", "r", stdin);
#endif
    int totDevice, vCnt = 0, totV = 0, maxV = 0, minV = 0;
    scanf("%d%d", &TotHoop, &totDevice);
    for (int i = 1; i <= TotHoop; i++)
        scanf("%d", i + Len);
    for (int i = 1; i <= totDevice; i++)
        scanf("%d", i + W);
    printf("%d
", DP2(TotHoop, totDevice, W, Len));
    return 0;
}
View Code
原文地址:https://www.cnblogs.com/headboy2002/p/8511849.html