Task Schedule

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13372 Accepted Submission(s): 4030

Problem Description
Our geometry princess XMM has stoped her study in computational geometry to concentrate on her newly opened factory. Her factory has introduced M new machines in order to process the coming N tasks. For the i-th task, the factory has to start processing it at or after day Si, process it for Pi days, and finish the task before or at day Ei. A machine can only work on one task at a time, and each task can be processed by at most one machine at a time. However, a task can be interrupted and processed on different machines on different days.
Now she wonders whether he has a feasible schedule to finish all the tasks in time. She turns to you for help.

Input
On the first line comes an integer T(T<=20), indicating the number of test cases.

You are given two integer N(N<=500) and M(M<=200) on the first line of each test case. Then on each of next N lines are three integers Pi, Si and Ei (1<=Pi, Si, Ei<=500), which have the meaning described in the description. It is guaranteed that in a feasible schedule every task that can be finished will be done before or at its end day.

Output
For each test case, print “Case x: ” first, where x is the case number. If there exists a feasible schedule to finish all the tasks, print “Yes”, otherwise print “No”.

Print a blank line after each test case.

Sample Input
2
4 3
1 3 5
1 1 4
2 3 7
3 5 9

2 2
2 1 3
1 2 2

Sample Output
Case 1: Yes

Case 2: Yes

一道比较裸的最大流。

看着挺像贪心的，如果不是网络流专题看到，应该会想贪心的做法。

考虑建图：因为任务比较少，我们把500天全部拿出来，让S连向每一天，流量为m，因为一共有m台机器，一天最多有m台机器同时进行。然后对于每个任务的st和ed，对于任务，让每一天连向这个任务，然后任务连向T，流量为p。

AC代码：

#pragma GCC optimize(2)
#include<bits/stdc++.h>
//#define int long long
using namespace std;
const int inf=0x3f3f3f3f;
const int N=10010,M=1000010;
int T,n,m,h[N],ts,s,t,res;
int head[N],nex[M],to[M],w[M],tot=1;
inline void ade(int a,int b,int c){
	to[++tot]=b; nex[tot]=head[a]; w[tot]=c; head[a]=tot;
}
inline void add(int a,int b,int c){
	ade(a,b,c);	ade(b,a,0);
}
inline int bfs(){
	memset(h,0,sizeof head);	queue<int> q;	q.push(s);	h[s]=1;
	while(q.size()){
		int u=q.front();	q.pop();
		for(int i=head[u];i;i=nex[i]){
			if(w[i]&&!h[to[i]]){
				h[to[i]]=h[u]+1;	q.push(to[i]);
			}
		}
	}
	return h[t];
}
int dfs(int x,int f){
	if(x==t)	return f;	int fl=0;
	for(int i=head[x];i&&f;i=nex[i]){
		if(w[i]&&h[to[i]]==h[x]+1){
			int mi=dfs(to[i],min(w[i],f));
			w[i]-=mi; w[i^1]+=mi; f-=mi; fl+=mi;
		}
	}
	if(!fl)	h[x]=-1;
	return fl;
}
inline int dinic(){
	int ret=0;
	while(bfs())	ret+=dfs(s,inf);
	return ret;
}
signed main(){
	cin>>T;
	while(T--){
		cin>>n>>m; s=res=0; t=n+501; tot=1; memset(head,0,sizeof head);
		for(int i=1;i<=500;i++)	add(s,i,m);
		for(int i=1;i<=n;i++){
			int s,p,e;	cin>>p>>s>>e;	res+=p;
			for(int j=s;j<=e;j++)	add(j,500+i,1);
			add(500+i,t,p);
		}
		printf("Case %d: ",++ts);
		puts(dinic()==res?"Yes":"No");puts("");
	}
	return 0;
}