hdu3572 Task Schedule (ISAP算法)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3572

题解：0为源点，源点与每个任务连一条边容量为天数pi,每个任务与相应时间点相连边容量为1，汇点dest=N+MAX+1,(其中max为结束时间的最大值，n为任务数）,每个时间点与汇点相连，边容量为m（机器数）。判断sum(pi)==maxflow.

#include <cstdio>    
#include <cstring>    
#define INF 0x3f3f3f3f   
#define MAXN 1200
#define MAXE 550000
using namespace std;  

struct node  
{  
	int to,cap,next;  
}edge[MAXE]; 

int n,src,dest,tot,pre[MAXN],cur[MAXN];    
int head[MAXN],level[MAXN],gap[MAXN],Queue[MAXN];  

void addEdge(int u,int v,int w)    
{    
	edge[tot].to=v;    
	edge[tot].cap=w;    
	edge[tot].next=head[u];    
	head[u]=tot++;    
	edge[tot].to=u;    
	edge[tot].cap=0;    
	edge[tot].next=head[v];    
	head[v]=tot++;    
}   

int ISAP()
{
	int v,u,minlevel,flag,flow,maxflow=0;
	memset(pre,-1,sizeof(pre));
	memset(level,0,sizeof(level));
	memset(gap,0,sizeof(gap));
	for(int i=0;i<=n;++i)
	{
		cur[i]=head[i];
	}
	u=pre[src]=src;
	flow=INF;
	gap[0]=n;
	while(level[src]<n)
	{
		flag=0;
		for(int &i=cur[u];i!=-1;i=edge[i].next)
		{
			v=edge[i].to;
			if(edge[i].cap>0&&level[u]==level[v]+1)
			{
				pre[v]=u;
				u=v;
				flag=1;
				flow=flow<edge[i].cap?flow:edge[i].cap;
				if(v==dest)
				{
					maxflow+=flow;
					for(u=pre[u];v!=src;v=u,u=pre[u])
					{
						edge[cur[u]].cap-=flow;
						edge[cur[u]^1].cap+=flow;
					}
					flow=INF;
				}
				break;
			}
		}
		if(flag)
			continue;
		minlevel=n;
		for(int i=head[u];i!=-1;i=edge[i].next)
		{
			v=edge[i].to;
			if(edge[i].cap>0&&minlevel>level[v])
			{
				cur[u]=i;
				minlevel=level[v];
			}
		}
		--gap[level[u]];
		if(gap[level[u]]==0)
			break;
		level[u]=minlevel+1;
		++gap[level[u]];
		u=pre[u];
	}
	return maxflow;
}

int main()
{
	int test,j,i,N,M,pi,si,ei,cases=0;
	int sum,MAX,ans;
	scanf("%d",&test);
	while(test--)
	{
		scanf("%d %d",&N,&M);
		tot=0,sum=0;
		MAX=0,src=0;
		memset(head,-1,sizeof(head));
		for(i=1;i<=N;++i)
		{
			scanf("%d %d %d",&pi,&si,&ei);
			sum+=pi;
			MAX=MAX>ei?MAX:ei;
			addEdge(src,i,pi);//源点到第i个任务连一条边
			for(j=si;j<=ei;++j)
				addEdge(i,N+j,1);//第i个任务到时间的连一条边(cap=1)
		}
		dest=N+MAX+1;//汇点
		n=dest+1;//点总数
		for(i=1;i<=MAX;++i)
		{
			addEdge(N+i,dest,M);//各时间点到汇点边cap=m
		}
		printf("Case %d: ",++cases);
		ans=ISAP();
		if(sum==ans)
			printf("Yes\n\n");
		else
			printf("No\n\n");
	}
	return 0;
}