POJ 1742 解题报告

来源：互联网发布：淘宝装修导航颜色代码编辑：程序博客网时间：2024/06/05 18:16

这道题是多重背包问题。我用的简单的二进制优化。主要是看到discuss里面这样就过了，也没去考虑别的优化。

代码基本上重用的POJ1276:http://blog.csdn.net/thestoryofsnow/article/details/41939829。

关于dp的类型，我本来用的是int，然后看dp[i] == i。这样超时了，看了discuss里面用bool，即dp[V] |= dp[V - cost]。这样就通过了。

thestoryofsnow1742Accepted228K2329MSC++1589B

/* ID: thestor1 LANG: C++ TASK: poj1742 */#include <iostream>#include <fstream>#include <cmath>#include <cstdio>#include <cstring>#include <limits>#include <string>#include <vector>#include <list>#include <set>#include <map>#include <queue>#include <stack>#include <algorithm>#include <cassert>using namespace std;const int MAXN = 100;const int MAXCASH = 100000;void zeroPack(int cost, bool dp[], const int cash){// zeroPackfor (int V = cash; V >= cost; --V){dp[V] |= dp[V - cost];}}void completePack(int cost, bool dp[], const int cash){for (int V = cost; V <= cash; ++V){dp[V] |= dp[V - cost];}}void multiplePack(int cnt, int cost, bool dp[], int cash){if (cnt * cost >= cash){// completePack// no constraints on item's 'cnt'completePack(cost, dp, cash);}else{// transform to finite items and applies zeroPackint k = 1;while (k < cnt){zeroPack(k * cost, dp, cash);cnt -= k;k <<= 1;}zeroPack(cnt * cost, dp, cash);}}int main(){int cnt[MAXN], cost[MAXN];bool dp[MAXCASH + 1];dp[0] = true;int n, m;while (scanf("%d%d", &n, &m) && !(n == 0 && m == 0)){for (int i = 0; i < n; ++i){scanf("%d", &cost[i]);}for (int i = 0; i < n; ++i){scanf("%d", &cnt[i]);}memset(dp + 1, false, m * sizeof(bool));for (int i = 0; i < n; ++i){multiplePack(cnt[i], cost[i], dp, m);}printf("%ld\n", count(dp + 1, dp + m + 1, true));}return 0;  }

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