2424.txt

北大ACM题目例程 详细的解答过程 程序实现 算法分析

//nclude"iostream.h"
#include"queue"
#include"stdio.h"
using namespace std;

struct node
{
	int time;
	int dinners;
};

int a[3],ans;
queue<node> have[3];


bool init()
{
	char c,t;
	int u[3],t1,t2,i;
	node nd,nd1;

	//cin>>a[0]>>a[1]>>a[2];
	scanf("%d%d%d",&a[0],&a[1],&a[2]);

	if(a[0]==0&&a[1]==0&&a[2]==0)
		return 0;
	
	for(i=0;i<3;i++)
	{
		while(!have[i].empty())
			have[i].pop();
	}
	
	ans=0;
	u[0]=0,u[1]=0,u[2]=0;

	while(1)
	{
		
		do{
			scanf("%c",&c);
		}while(! (c=='#'||(c<='9'&&c>='0')) );
	
		if(c=='#')break;
	
		scanf("%d:%d %d",&t1,&t2,&nd.dinners);

		nd.time=((c-'0')*10+t1)*60+t2;
		i=(nd.dinners-1)/2;
		
	//	if(i>=3||i<0)continue;


		while(!have[i].empty())
		{
			nd1=have[i].front();
				
			if(nd1.time>nd.time)
				break;
			
			ans+=nd1.dinners;
			have[i].pop();
			u[i]--;
		}

		if(u[i]==a[i]&&!have[i].empty())
		{
			nd1=have[i].front();
			if(nd1.time<=nd.time+30)
			{
				ans+=nd1.dinners;
				nd.time=nd1.time;

				have[i].pop();
				u[i]--;
			}
		}

		if(u[i]<a[i])
		{
			u[i]++;
			nd.time+=30;
			if(nd.time<=23*60)
				have[i].push(nd);
		}
	}
	
	for(i=0;i<3;i++)
	while(!have[i].empty())
	{
		nd1=have[i].front();
		ans+=nd1.dinners;
		have[i].pop();
		u[i]--;
	}
	
	return 1;

}	

int main()
{
	while(init())
	{
		printf("%d\n",ans);
	}
	return 0;
}