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 Mstrings 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
http://blog.csdn.net/lyhvoyage/article/details/17041907
//* @author:
import java.util.*;
class Cloth{//表示一件要洗的衣服
int time;//洗这件衣服的时间
String color;//这件衣服的颜色
public Cloth(){
this.time=0;
this.color=null;
}
public Cloth(int tiem,String color){
this.time=time;
this.color=color;
}
}
public class Main{
private int n;//颜色种类数
private int m;//衣服件数
private String colors[];//颜色数组，保存所有的颜色
private Cloth cloths[];//衣服数组，保存所有要洗的衣服
private int res;//洗完全部衣服所需的最少时间
public Main(int n,int m,String colors[],Cloth cloths[]){
this.n=n;
this.m=m;
this.colors=colors;
this.cloths=cloths;
res=0;
}
public void dp(){
int r[]=new int[n];//r[j]表示一个人洗colors[j]这种颜色的衣服的总的洗衣时间
int dp[]=new int[50000];
/*
dp[j]表示在时间j内, 一个人洗某种颜色的衣服，一件件的洗. 最大洗衣时间。
总的衣服数< 100,每件衣服的洗衣时间< 1000，100*1000/2=50000
*/
for(int i=0;i< m;i++){//对每一件衣服
for(int j=0;j< n;j++){//看它是哪一种的颜色
if(cloths[i].color.equals(colors[j])){
r[j]+=cloths[i].time;//计算这种颜色的衣服总的洗衣时间
break;
}
}
}
for(int i=0;i< n;i++){//对每一种颜色（n个01背包问题）
for(int j=0;j<=r[i]/2;j++)
dp[j]=0;
for(int j=0;j< m;j++){//在所有衣服中找这种颜色的衣服
if(cloths[j].color.equals(colors[i])){
for(int k=r[i]/2;k>=cloths[j].time;k--){
if(dp[k]< dp[k-cloths[j].time]+cloths[j].time)
dp[k]=dp[k-cloths[j].time]+cloths[j].time;
}
}
}
res+=r[i]-dp[r[i]/2];//洗完全部衣服所需的最少时间=洗各种颜色衣服所需最少时间的和。
}
System.out.println(res);
}
public static void main(String args[]){
String colors[];
Cloth cloths[];
Scanner in=new Scanner(System.in);
while(true){
int n=in.nextInt();
int m=in.nextInt();
if(n==0&&m==0) break;
colors=new String[n];
cloths=new Cloth[m];
for(int i=0;i< n;i++)
colors[i]=in.next();
for(int i=0;i< m;i++){
cloths[i]=new Cloth();
cloths[i].time=in.nextInt();
cloths[i].color=in.next();
}
Main mm=new Main(n,m,colors,cloths);
mm.dp();
}
}
}