01背包的变形

http://poj.org/problem?id=3211

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

/**
poj3211
题目大意:夫妻两个人一起洗衣服，洗衣速度一样快，衣服有多种颜色，要求只有在完全洗完一种颜色的衣服后才能洗另一种颜色的，问所有的衣服都洗完最少用时是多少？
解题思路：把相同颜色的衣服看做一类，统计总时间sum然后将sum尽可能的均分的分给两个人（借助01背包，详见代码），取最大值加入ans。最后所有一份都统计完，ans即为答案。
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
struct node
{
int ip;
char a[25];
}p[105];
bool cmp(node x,node y)
{
return strcmp(x.a,y.a)>0;
}
int n,m;
int s[1005],dp[100005];
int main()
{
while(~scanf("%d%d%*c",&m,&n))
{
if(n==0&&m==0)
break;
char b[1005];
gets(b);
for(int i=0;i<n;i++)
scanf("%d%s",&p[i].ip,p[i].a);
p
.ip=0;
strcpy(p[n++].a,"####");
sort(p,p+n,cmp);
/* for(int i=0;i<n;i++)
printf("%d %s\n",p[i].ip,p[i].a);*/
int sum=0,k=1,ans=0;
for(int i=0;i<n-1;i++)
{
if(strcmp(p[i].a,p[i+1].a)==0)
{
s[k++]=p[i].ip;
sum+=s[k-1];
}
else
{
s[k]=p[i].ip;
sum+=s[k];
memset(dp,0,sizeof(dp));
int cost=sum/2;
for(int j=1;j<=k;j++)
{
for(int l=cost;l>=s[j];l--)
dp[l]=max(dp[l],dp[l-s[j]]+s[j]);
}
ans+=max(dp[cost],sum-dp[cost]);
sum=0;
k=1;
}
}
printf("%d\n",ans);
}
return 0;
}

/**
3 1
red blue yellow
2 red
*/