ZOJ3116 POJ3483 HDU1910 Loan Scheduling

没有任何难度的贪心。先按照时间预排序，对于每一新的任务，先判定是否还有处理的空间，如果没有就往前检索看看能不能够踢掉某些受益少的任务。

/*******************************************************************************
 # Author : Neo Fung
 # Email : [email protected]
 # Last modified: 2012-01-12 19:22
 # Filename: ZOJ3116 POJ3483 HDU1910 Loan Scheduling.cpp
 # Description : 
 ******************************************************************************/
#ifdef _MSC_VER
#define DEBUG
#endif

#include <fstream>
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <string>
#include <limits.h>
#include <algorithm>
#include <math.h>
#include <numeric>
#include <functional>
#include <ctype.h>
#define MAX 10010
using namespace std;

pair<int ,int> app[MAX];
bool used[MAX];

bool inline cmp(const pair<int,int> &lhs,const pair<int,int> &rhs)
{
	if(lhs.first==rhs.first)
		return lhs.second>rhs.second;
	else
		return lhs.first<rhs.first;
}

int main(void)
{
#ifdef DEBUG  
  freopen("../stdin.txt","r",stdin);
  freopen("../stdout.txt","w",stdout); 
#endif  
	int n,l;

	while(~scanf("%d%d",&n,&l))
	{
		for(int i=0;i<n;++i)
			scanf("%d%d",&app[i].second,&app[i].first);
		sort(app,app+n,cmp);
		memset(used,false,sizeof(used));

		int remain=l,now=0;
		int ans=0;
		for(int i=0;i<n;++i)
		{
			remain +=(app[i].first-now)*l;
			if(now<=app[i].first && remain)
			{
				--remain;
				ans+=app[i].second;
				used[i]=true;
			}
			else if(now<=app[i].first && !remain)
			{
				int t_ind=i;
				for(int j=i-1;j>=0;--j)
					if(used[j] && app[j].second<app[t_ind].second)
						t_ind=j;
				if(t_ind!=i)
				{
					used[t_ind]=false;
					ans-=app[t_ind].second;
					used[i]=true;
					ans+=app[i].second;
				}
			}
			now = app[i].first;
		}
		printf("%d\n",ans);
	}

  return 0;
}