poj3211 Washing Clothes(01 가방 문제)

Washing Clothes
Time Limit: 1000MS
 
Memory Limit: 131072K
Total Submissions: 6490
 
Accepted: 1828
Description
Dearboy was so busy recently that now he has piles of clothes to wash. Luckily, he has a beautiful and hard-working girlfriend to help him. The clothes are in varieties of colors but each piece of them can be seen as of only one color. In order to prevent the clothes from getting dyed in mixed colors, Dearboy and his girlfriend have to finish washing all clothes of one color before going on to those of another color.
From experience Dearboy knows how long each piece of clothes takes one person to wash. Each piece will be washed by either Dearboy or his girlfriend but not both of them. The couple can wash two pieces simultaneously. What is the shortest possible time they need to finish the job?
Input
The input contains several test cases. Each test case begins with a line of two positive integers M and N (M < 10, N < 100), which are the numbers of colors and of clothes. The next line contains M strings which are not longer than 10 characters and do not contain spaces, which the names of the colors. Then follow N lines describing the clothes. Each of these lines contains the time to wash some piece of the clothes (less than 1,000) and its color. Two zeroes follow the last test case.
Output
For each test case output on a separate line the time the couple needs for washing.
Sample Input

3 4
red blue yellow
2 red
3 blue
4 blue
6 red
0 0

Sample Output

10

Source
POJ Monthly--2007.04.01, dearboy
제목:http://poj.org/problem?id=3211
분석: 이 문제는 조금만 더 숨기면 01배낭 문제입니다. 배낭을 채워야 할 수도 있고 가장 많이 담아야 할 상황으로 바뀔 수도 있습니다. 즉, 01배낭으로 몇 개의 수를 판단하고 조합을 통해 이 수의 와/2에 최대한 접근할 수 있습니다.아주 좌절 했 다 = = 구체적으로 코드 를 보자!
코드:

#include<cstdio>
#include<cstring>
using namespace std;
const int mm=111;
int g[mm][mm],t[mm],s[mm];
bool f[111111];
char color[mm][22],tmp[22];
int n,m;
int getid(char *s)
{
    for(int i=0;i<m;++i)
        if(strcmp(s,color[i])==0)return i;
    return 0;
}
int main()
{
    int i,j,k,ans;
    while(scanf("%d%d",&m,&n),n+m)
    {
        for(i=0;i<m;++i)scanf("%s",color[i]),t[i]=s[i]=0;
        for(i=0;i<n;++i)
        {
            scanf("%d%s",&k,tmp);
            s[j=getid(tmp)]+=k;
            g[j][t[j]++]=k;
        }
        for(ans=k=0;k<m;++k)
        {
            for(i=0;i<=s[k]/2;++i)f[i]=0;
            f[0]=1;
            for(i=0;i<t[k];++i)
                for(j=s[k]/2-g[k][i];j>=0;--j)
                    if(f[j])f[j+g[k][i]]=1;
            for(i=s[k]/2;i>=0;--i)
                if(f[i])break;
            ans+=s[k]-i;
        }
        printf("%d
",ans);
    }
    return 0;
}