kebab

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Almost everyone likes kebabs nowadays (Here a kebab means pieces of meat grilled on a long thin stick). Have you, however, considered about the hardship of a kebab roaster while enjoying the delicious food?

Well, here‘s a chance for
you to help the poor roaster make sure whether he can deal with the following orders without dissatisfying the customers.

Now N customers is coming. Customer i will arrive at time si (which means the roaster cannot serve customer i until time si). He/She will order ni kebabs, each one of which requires a total amount of ti unit time to get it well-roasted, and want to get them
before time ei(Just at exactly time ei is also OK). The roaster has a big grill which can hold an unlimited amount of kebabs (Unbelievable huh? Trust me, it’s real!). But he has so little charcoal that at most M kebabs can be roasted at the same time. He is
skillful enough to take no time changing the kebabs being roasted. Can you help him determine if he can meet all the customers’ demand?

Oh, I forgot to say that the roaster needs not to roast a single kebab in a successive period of time. That means he can divide the whole ti unit time into k (1<=k<=ti) parts such that any two adjacent parts don’t have to be successive in time. He can also
divide a single kebab into k (1<=k<=ti) parts and roast them simultaneously. The time needed to roast one part of the kebab well is linear to the amount of meat it contains. So if a kebab needs 10 unit time to roast well, he can divide it into 10 parts and
roast them simultaneously just one unit time. Remember, however, a single unit time is indivisible and the kebab can only be divided into such parts that each needs an integral unit time to roast well.

Input

There are multiple test cases. The first line of each case contains two positive integers N and M. N is the number of customers and M is the maximum kebabs the grill can roast at the same time. Then follow N lines each describing
one customer, containing four integers: si (arrival time), ni (demand for kebabs), ei (deadline) and ti (time needed for roasting one kebab well).

There is a blank line after each input block.

Restriction:

1 <= N <= 200, 1 <= M <= 1,000

1 <= ni, ti <= 50

1 <= si < ei <= 1,000,000

Output

If the roaster can satisfy all the customers, output “Yes” (without quotes). Otherwise, output “No”.

Sample Input

2 10
1 10 6 3
2 10 4 2

2 10
1 10 5 3
2 10 4 2

Sample Output

Yes
No

Source

2009 Multi-University Training Contest 9 - Host by HIT

题意描写叙述：有n个人来烤肉店吃烤肉。每一个人在si 时刻来ei 时刻离开而且点了ni 份，

每份烤肉要烤到ti 个单位时间才算烤熟，烤肉店里能够同一时候烤m份。问是否有一种计划

使得n个人都能够拿到自己的ni 份。

參考大牛解题思路：这道题本身不是非常难，网络流的模型也非经常见，可是这道题中（si，ei）的时间

跨度非常大（1<=si<=ei<=1000000），所以不能把时间区间直接拆分开建立模型。这样顶点

个数太多，会超时。这里，介绍一下学到的新技巧，我们能够把时间区间压缩：

time[]里保存所有的si 和 ei ，这样time[i]-time[i-1]就表示一段时间区间了。

这题和HDU 3572相似，但又不能像那题那样做，由于这题时间长度有点大

所以将时间区间当成一个点。将该区间连向超级汇点，容量为区间长度*M

将全部客人连向超级源点。容量为烤肉数量＊每串烤肉所需时间

接下来的难点就是怎么将客人和时间区间连起来了 ,

假设时间区间在客人来的时间和走的时间这段区间内，

就表明这段时间能够用来帮客人烤肉，所以能够连接。容量为inf

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<queue>
using namespace std;
#define M 3000
#define inf 0x3f3f3f3f
int head[M],dis[M],st,t,n,m,cnt;
struct node{
	int v,next,w;
}mp[M*M];
void add(int u,int v,int w){
	mp[cnt].v=v;
	mp[cnt].w=w;
	mp[cnt].next=head[u];
	head[u]=cnt++;
	mp[cnt].v=u;
	mp[cnt].w=0;
	mp[cnt].next=head[v];
	head[v]=cnt++;
}
int bfs(){
	memset(dis,-1,sizeof(dis));
	queue <int> q;
	dis[st]=0;
	q.push(st);
	while(!q.empty()){
		int u=q.front();
		q.pop();
		for(int i=head[u];i!=-1;i=mp[i].next){
			int v=mp[i].v;
			if( mp[i].w>0 && dis[v]==-1){
				dis[v]=dis[u]+1;
				if(v==t) return 1;
				q.push(v);
			}
		}
	}
	return 0;
}
int dinic(int s,int low){//依照凝视地方写就会wa，曾经这么写就能过啊。这次快wa哭了，，= =+
	if(s==t || low==0) return low;
	int a,ans=low;//ans=0;
	for(int i=head[s];i!=-1;i=mp[i].next){
		int v=mp[i].v;
		if(mp[i].w>0 && dis[v]==dis[s]+1 && (a=dinic(v,min(ans/*low*/,mp[i].w)))){
			mp[i].w-=a;
			mp[i^1].w+=a;
		//	ans+=a;
		//	if(ans==low) break;
			ans-=a;
			if(ans==0) return low;
		}
	}
	//return ans;
	return low-ans;
}
int main(){
	int tot,count,sum;
	int s[M],e[M],num[M],ti[M],time[M];
	while(~scanf("%d%d",&n,&m)){
		sum=cnt=0;tot=1; count=0;
		memset(head,-1,sizeof(head));
		memset(time,0,sizeof(time));
		for(int i=1;i<=n;i++){
			scanf("%d%d%d%d",&s[i],&num[i],&e[i],&ti[i]);
			sum+=num[i]*ti[i];
			time[tot++]=s[i];
			time[tot++]=e[i];
		}
		sort(time+1,time+tot);
		for(int i=1;i<tot;i++)//消除反复区域
			if(time[count]!=time[i])
				time[++count]=time[i];
		st=n+count+1;//起点
		t=st+1; //汇点
		for(int i=1;i<=n;i++)//起点到每一个顾客。权值为烤肉数乘以时间
			add(st,i,num[i]*ti[i]);
		for(int i=1;i<=count;i++){
			add(n+i,t,m*(time[i]-time[i-1]));//时间区间到汇点，权值为单位时间完毕烤肉m乘以区间长度
			for(int j=1;j<=n;j++){
				if(s[j]<=time[i-1]&&e[j]>=time[i])
					add(j,n+i,inf);//假设顾客的区间段
			}
		}
		int ans=0;
		while(bfs())
			ans+=dinic(st,inf);
		if(sum==ans) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}