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
额，01背包，开始以为贪心，我还是太年轻了不说了。策略如下：
统计每种颜色衣服需要的总时间sum,然后sum/2进行01背包，看能够装下的最大时间
具体见挫码=。=
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
using namespace std;
int num[15][110];//统计每种颜色衣服各个数量所需时间
int nu[15];//统计每种颜色衣服的数量
char str[15][15];//读入字符
int dp[10000];//背包，开始开下了，wa了=。=
int n,m;
int main()
{
int t;
char s[110];
while(~scanf("%d%d",&n,&m)&&(n+m))
{
getchar();
for(int i=0;i<n;i++)
{
scanf("%s",str[i]);
nu[i]=0;
}
for(int i=0;i<m;i++)
{
scanf("%d%s",&t,s);
for(int j=0;j<n;j++)
{
if(!strcmp(str[j],s))
{
nu[j]++;
num[j][nu[j]-1]=t;
}
}
}
int ans=0;
for(int i=0;i<n;i++)
{
if(nu[i])
{
int sum=0;
for(int j=0;j<nu[i];j++)
sum+=num[i][j];
memset(dp,0,sizeof(dp));//01背包枚举每种物品
for(int j=0;j<nu[i];j++)
{
for(int k=sum/2;k>=num[i][j];k--)
dp[k]=max(dp[k],dp[k-num[i][j]]+num[i][j]);
}
ans+=max(dp[sum/2],sum-dp[sum/2]);
}
}
printf("%d\n",ans);
}
return 0;
}
POJ 3211 Washing Clothes