BOJ 385

Problem C. Vaccine

Description

On Christmas Eve, 2029, the unidentified "Apocalypse Virus" spreads and

plunges JA into a state of emergency in a chaos known as the "Lost

Christmas" incident. An international organization known as the GHQ

intervenes with martial law and restores order to JA at the cost of its

independence. GHQ takes the domination of JA by the monopolization of

the vaccines, which are the only known solution to prevent the spreading

of the fatal Apocalypse Virus.

However, where there is oppression, there is resistance. The armed NGO

“Undertakers” is such a resistance group that aims to liberate JA from the

GHQ. One of the members' task is to manufacture the Anti-Apocalypse-

Virus vaccines. Of course, the government GHQ won't give them the raw

material as the Undertakers is announced as a terroristic organization. So

they have to get the material on their own.

Just as the sources of the raw material they get are various, the material

should be purified. The process of purification for each portion of raw

material needs 2 steps IN THE FOLLOWING ORDER:

1. take a portion of raw material into machine A to heat it to

142.857℃.

2. take the portion of raw material into machine B to centrifugalize it .

And the problem comes: Undertakers has only one machine A and only

one machine B. The machine A is able to handle only one portion of

material at a time, so is the machine B. However, there are N portions of

raw material in all, and the time for heating them and for centrifugalizing

them differs. So how to make a proper plan to purify them to minimumize

the total time? You may assume that taking a portion into a machine or

taking a portion out from a machine cost no time.

Input

There are multiple cases, end by EOF.

For each case, the first line contains an integer N (1 <= N <= 1000), the

number of portions of raw material. Then N lines followed, each contains

two integers, time for heat the ith portion of material and for

centrifugalize it respectively.

Output

For each case, output one integer, the minimum total time for purifying

these portions of raw material.

Sample Input

2 3

5

4 3

5 8

5

9 2

4 10

6 8

3 5

Sample Output

11

33

Hint

Take the first example for explaining. The following process takes the

minimum time:

At the beginning, take the first portion of material into machine A.

At the 3rd minute, take out the first portion from machine A and take it

into machine B, and then take the second portion into machine A.

At the 7th minute, take out the second portion from machine A.

At the 8th minute, take out the first portion from machine B and take the

second portion into machine B.

At the 11th minute, take out the second portion from machine B

贪心的思路，利用排序不等式排序，然后求总时间。

#include<stdio.h>
#include<stdlib.h>
struct raw{
	int a,b;
}lis[1010];
int max(int a,int b){
	return a>b?a:b;
}
int cmp(const void * a,const void *b){
	struct raw *aa=(struct raw *)a;
	struct raw *bb=(struct raw *)b;
	if(aa->a+bb->b+max(aa->b,bb->a)>aa->b+bb->a+max(aa->a,bb->b))
		return 1;
	return 0;
}
int main(){
	int i,j,n;
	while(scanf("%d",&n)!=EOF){
		for(i=1;i<=n;i++){
			scanf("%d %d",&lis[i].a,&lis[i].b);
		}
		qsort(&lis[1],n,sizeof(lis[1]),cmp);
		int s=0,t=0;       //这里s记录上一过程中A机器运行结束时间，t记录上一过程中B机器运行结束时间
		for(i=1;i<=n;i++){
			if(s+lis[i].a>t)
				t=s+lis[i].a;
			s+=lis[i].a;
			t+=lis[i].b;
		}
		printf("%d\n",t);
	}
}