HDU-3732 Ahui Writes Word 解题报告

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 20

go 5 8

think 3 7

big 7 4

read 2 6

write 3 5

Sample Output

15

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3732

解法类型：贪心+01背包 || 多重背包

解题思路：乍一看是很典型的01背包问题，但一看数据，01背包的复杂度O(N*V)对于这题来说肯定会超时，而且01背包要优化也只能是常数级。我们注意到01背包方程:d[v]=max{d[v],d[v-cost]+weight};d[v]存储的是前i个物品在最大容量V下所能获得的最大价值。再仔细看题目，注意这个条件0
≤ Vi , Ci ≤ 10。由于vi和ci都在很小的范围内，所以可以用贪心的思想先求出当单词复杂度不超过C-10时的最大价值，假设此时为Cmax，这样D[Cmax]是最大的。然后用01背包求出剩下的容量Cmax-C所能取得的最大价值，最后相加即可。显然，时间复杂度几乎是线性的！

当然这题也可以优化为多重背包来求解，同样注意0
≤ Vi , Ci ≤ 10，所以N最大为100，时间复杂度也不是很高。

算法实现：

贪心+01背包：

//STATUS:C++_AC_187MS_1008KB
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
const int MAX=100010;
int cmp(const void* a,const void* b);
struct NUM
{
    int v,c;
}num[MAX];
int N,C,d[20];
int main()
{
//    freopen("in.txt","r",stdin);
    int i,ansv,ansc,lc,c,t;
    char ch[20];
    while(~scanf("%d%d",&N,&C))
    {
        memset(d,0,sizeof(d));
        for(i=0;i<N;i++)
            scanf("%s%d%d",ch,&num[i].v,&num[i].c);
        qsort(num,N,sizeof(NUM),cmp);

        ansv=ansc=0;
        for(i=0;i<N;i++){
            ansv+=num[i].v;
            ansc+=num[i].c;
            if(C-ansc<10){
                ansv-=num[i].v;
                ansc-=num[i].c;
                break;                
            }
        }
        if(i<N){
            for(lc=C-ansc;i<N;i++){
                for(c=lc;c>=num[i].c;c--){
                    t=d[c-num[i].c]+num[i].v;
                    if(t>d[c])d[c]=t;
                }
            }
            ansv+=d[lc];
        }
        printf("%d\n",ansv);
    }
    return 0;
}

int cmp(const void* a,const void* b)
{
    int v1=((NUM*)a)->v,c1=((NUM*)a)->c,
        v2=((NUM*)b)->v,c2=((NUM*)b)->c;
    if(v1*c2>v2*c1)return -1;
    else if(v1*c2<v2*c1)return 1;
    else return 0;
}

多重背包：

//STATUS:C++_AC_171MS_268KB
#include<stdio.h>
#include<string.h>
const int MAX1=10010,MAX2=110;
void cpack(int cost,int weight);
void zopack(int cost,int weight);
int map[11][11],d[MAX1],N,C,vc[MAX2][2],n[MAX2],nup[MAX2];
int main()
{
//    freopen("in.txt","r",stdin);
    int i,k,v,c,t;
    char ch[20];
    while(~scanf("%d%d",&N,&C))
    {
        memset(map,0,sizeof(map));
        memset(d,0,sizeof(d));
        for(i=0;i<N;i++){
            scanf("%s%d%d",ch,&v,&c);
            map[v][c]++;
        }

        for(k=0,v=1;v<11;v++)
            for(c=1;c<11;c++)
                if(map[v][c])
                    vc[k][0]=v,vc[k][1]=c,
                    n[k]=map[v][c],nup[k++]=C/c;
        N=k;
        for(i=0;i<N;i++){
            if(n[i]>=nup[i])
                cpack(vc[i][1],vc[i][0]);
            else {
                t=n[i];
                for(k=1;k<t;){
                    zopack(k*vc[i][1],k*vc[i][0]);
                    t-=k;
                    k*=2;
                }
                zopack(t*vc[i][1],t*vc[i][0]);
            }
        }
        
        printf("%d\n",d[C]);
    }
    return 0;
}

void cpack(int cost,int weight)
{
    int c,t;
    for(c=cost;c<=C;c++)
        if((t=d[c-cost]+weight)>d[c])d[c]=t;
}
void zopack(int cost,int weight)
{
    int c,t;
    for(c=C;c>=cost;c--)
        if((t=d[c-cost]+weight)>d[c])d[c]=t;
}