多重背包

Ahui Writes Word

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 908 Accepted Submission(s): 269 

Problem Description

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.

 

Input

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

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

 

Sample Input

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

 

Sample Output

15
Hint
Input data is huge,please use “scanf(“%s”,s)”
#include <iostream>#include <sstream>#include <string.h>#include <cstdio>#include <string>#include <cctype>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <set>#include <cmath>#define ll long long#define INF 0x7fffffff#define MV 1005#define ME 10005#define N 100005using namespace std;int in[12][12];int dp[N];int w[N];int v[N];int main(){    int W;    int n;    while (~scanf("%d%d", &n, &W)) {        char s[12];        memset(in, 0, sizeof in);        for (int i = 0; i < n; i++) {            int v, w;            scanf("%s%d%d", s, &v, &w);            in[v][w]++;        }        /// 转化成01被背包        /// 二进制拆分 13 = 1 + 2 + 4 + 6        int num = 0;        for (int i = 0; i <= 10; i++) {            for (int j = 0; j <= 10; j++) {               // if (in[i][j] > 0) {                    int temp = in[i][j];                    for (int k = 1; k <= temp; k = k*2) {                        v[num] = i*k;                        w[num] = j*k;                        num++;                        temp -= k;                    }                    if (temp > 0) {                        v[num] = i*temp;                        w[num] = j*temp;                        num++;                    }                //}            }        }        memset(dp, 0, sizeof dp);        for (int i = 0; i < num; i++) {            for (int j = W; j >= w[i]; j--) {                dp[j] = max(dp[j], dp[j-w[i]] + v[i]);            }        }        printf("%d\n", dp[W]);    }    return 0;}