zoukankan      html  css  js  c++  java
  • lightoj1233_二进制优化

    http://lightoj.com/volume_showproblem.php?problem=1233

    一维多重背包二进制优化的思考:先看这个问题,100=1+2+4+8+16+32+37,观察可以得出100以内任何一个数都可以由以上7个数选择组合得到,所以对物品数目就不是从0都100遍历,而是0,1,2,4,5,16,32,37遍历,时间大大优化。

    In a strange shop there are n types of coins of value A1, A2 ... AnC1, C2, ... Cn denote the number of coins of value A1, A2 ... An respectively. You have to find the number of different values (from 1 to m), which can be produced using these coins.

    Input

    Input starts with an integer T (≤ 20), denoting the number of test cases.

    Each case starts with a line containing two integers n (1 ≤ n ≤ 100), m (0 ≤ m ≤ 105). The next line contains 2n integers, denoting A1, A2 ... An, C1, C2 ... Cn (1 ≤ Ai ≤ 105, 1 ≤ Ci ≤ 1000). All Ai will be distinct.

    Output

    For each case, print the case number and the result.

    Sample Input

    Output for Sample Input

    2

    3 10

    1 2 4 2 1 1

    2 5

    1 4 2 1

    Case 1: 8

    Case 2: 4

     1 #include <algorithm>
     2 #include <iostream>
     3 #include <cstring>
     4 #include <cstdlib>
     5 #include <cstdio>
     6 #include <vector>
     7 #include <ctime>
     8 #include <queue>
     9 #include <list>
    10 #include <set>
    11 #include <map>
    12 using namespace std;
    13 #define INF 0x3f3f3f3f
    14 typedef long long LL;
    15 const int N = 1e5 + 5;
    16 int dp[N];
    17 int val[105], c[105];
    18 
    19 int main()
    20 {
    21     int t, n, m;
    22     scanf("%d", &t);
    23     for(int ca = 1; ca <= t; ++ca) {
    24         scanf("%d %d", &n, &m);
    25         for(int i = 1; i <= n; ++i) {
    26             scanf("%d", val + i);
    27         }
    28         for(int i = 1; i <= n; ++i) {
    29             scanf("%d", c + i);
    30         }
    31         memset(dp, 0, sizeof(dp));
    32         dp[0] = 1;
    33         for(int i = 1; i <= n; ++i) {
    34             for(int k = 1; k <= c[i]; k <<= 1) {
    35                 for(int j = m; j >= k*val[i]; --j) {
    36                     dp[j] |= dp[j - k*val[i]];
    37                 }
    38                 c[i] -= k;
    39             }
    40             if(c[i]) {
    41                 for(int j = m; j >= c[i]*val[i]; --j) {
    42                     dp[j] |= dp[j - c[i]*val[i]];
    43                 }
    44             }
    45         }
    46         int ans = 0;
    47         for(int i = 1; i <= m; ++i) {
    48             ans += dp[i];
    49         }
    50         printf("Case %d: %d
    ", ca, ans);
    51     }
    52     return 0;
    53 }
  • 相关阅读:
    Selenium+PhantomJS实现简易有道翻译爬虫
    Scrapy框架实战-妹子图爬虫
    拉勾网职位信息爬取
    Docker Compose容器编排
    Ansible进阶--playbook的使用
    etcd集群部署
    使用Dockerfile构建镜像
    Docker网络管理
    Docker数据管理
    Dubbo高性能网关--Flurry介绍
  • 原文地址:https://www.cnblogs.com/luomi/p/5965163.html
Copyright © 2011-2022 走看看