学步园

最大流问题：dinic做的，isap正在学习中

建图还是比较容易的：

源点与每个任务建边流量为任务需要的时间，任务与时间段的每个时间点建边流量为1，

每个时间点与汇点建边流量为机器的个数

#include<stdio.h>
#include<queue>
#include<string.h>
#define N 3000
#define inf 0x3fffffff
using namespace std;
int head[N],num,d[N],n,et;
struct edge
{
	int st,ed,flow,next;
}E[500100];
void addedge(int x,int y,int w)
{
	E[num].st=x;
	E[num].ed=y;
	E[num].flow=w;
	E[num].next=head[x];
	head[x]=num++;
}
bool bfs(int sp,int tp)
{
	int u,v,i;
	memset(d,-1,sizeof(d));
	queue<int>Q;
	d[sp]=0;
	Q.push(sp);
	while(!Q.empty())
	{
		u=Q.front();
		Q.pop();
		for(i=head[u];i!=-1;i=E[i].next)
		{
			v=E[i].ed;
			if(d[v]==-1&&E[i].flow>0)
			{
				d[v]=d[u]+1;
				Q.push(v);
			}
		}
	}
	if(d[tp]==-1)return false;
	return true;
}
int dfs(int u,int minflow)
{
	if(u==et)return minflow;
	int i,v,r=0,f;
	for(i=head[u];i!=-1;i=E[i].next)
	{
		v=E[i].ed;
		if(d[v]==d[u]+1&&E[i].flow>0)
		{
			v=E[i].ed;
			if(d[v]==d[u]+1&&E[i].flow>0&&r<minflow&&(f=dfs(v,E[i].flow>minflow-r?minflow-r:E[i].flow)))
			{
				E[i].flow-=f;
				E[i^1].flow+=f;
				r+=f;
			}
		}		
	}
	if(f==0)d[u]=-1;
		return r;
}
int dinic(int sp,int tp)
{
	int maxflow=0,sum;
	while(bfs(sp,tp))
	{
		while(sum=dfs(sp,inf))
		 maxflow+=sum;
	}
	return maxflow;
}
int main()
{
	int a,b,c,i,m,t,op=1,j,max,sum,ans;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		memset(head,-1,sizeof(head));
		num=0;max=0;sum=0;
		for(i=1;i<=n;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			sum+=a;
			addedge(0,i,a);
			addedge(i,0,0);
			if(max<c)max=c;
			for(j=b;j<=c;j++)
			{
              addedge(i,n+j,1);
			  addedge(n+j,i,0);
			}
		}
		et=n+max+1;
		for(i=1;i<=max;i++)
		{
			   addedge(n+i,et,m);
			   addedge(et,n+i,0);
		}
        ans=dinic(0,et);
		printf("Case %d: ",op++);
		if(sum==ans)
		printf("Yes\n");
		else printf("No\n");
		printf("\n");
	}
	return 0;
}