HDOJ 3732 Ahui Writes Word

We all know that English is very important, so Ahui strive for this in order to learn more English words. To know that word has its value and complexity of writing (the length of each word does not exceed 10 by only lowercase letters), Ahui wrote the complexity of the total is less than or equal to C.
Question: the maximum value Ahui can get.
Note: input words will not be the same.

The first line of each test case are two integer N , C, representing the number of Ahui’s words and the total complexity of written words. (1 ≤ N ≤ 100000, 1 ≤ C ≤ 10000)
Each of the next N line are a string and two integer, representing the word, the value(Vi ) and the complexity(Ci ). (0 ≤ Vi , Ci ≤ 10)

Output the maximum value in a single line for each test case.

  
5 20
go 5 8
think 3 7
big 7 4
read 2 6
write 3 5

  
15
Hint
Input data is huge,please use “scanf(“%s”,s)”
题目分析：
首先，我们可以明确一点，输入数据中的字符串是没有作用的。然后，还有一点要注意，这道题目乍看上去像是01背包（注意数据量N≤ 100000，C ≤ 10000，直接套用01背包显然会超时），但是实际上是多重背包（0 ≤ Vi , Ci ≤ 10, (Vi,Ci)的组合最多就是121个，但是N最大是100000，所以里面会出现很多重复的（Vi，Ci），也就是多重背包了）。
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define max(x,y) x>y?x:yint dp[10010], num[12][12], weight[100010], value[100010];int main()
{int n, c;while(~scanf("%d%d", &n, &c)){int valuea, price, count = 0;char s[20];memset(dp, 0, sizeof(dp));memset(num, 0, sizeof(num));memset(weight, 0, sizeof(weight));memset(value, 0, sizeof(value));for(int i = 1; i <= n; i++){scanf("%s%d%d", s, &valuea, &price);num[valuea][price]++;}for(int i = 0; i <= 10; i++){for(int k = 0; k <= 10; k++){int tmp = num[i][k];for(int j = 1; j <= tmp; j = j*2) // 二进制拆分         {            weight[count] = j * k;            value[count++] = j * i;             tmp -= j;        }if(tmp > 0)        {            weight[count] = tmp * k;             value[count++] = tmp * i;         } }}for(int i = 0; i < count; i++){for(int j = c; j >= weight[i]; j--){dp[j] = max(dp[j], dp[j-weight[i]]+value[i]);}    }printf("%d\n", dp[c]);}return 0;
}