hdu 3572 Task Schedule hdu 2883 kebab 最大流

注意一个地方,在建立好层次图进行dfs增广时,如果某个点增广不出流量,那么将该点level值置为0,即将它在该层次图中删除,剪枝,否则会超时~

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=1050;
#define inf 10000000
int n,m;
int level[maxn],que[maxn*5];
int head[maxn],lon;
int min(int a,int b)
{
	if(a<b) return a;
	else return b;
}
struct
{
    int next,to,c;
}e[500100];
void edgeini()
{
    memset(head,-1,sizeof(head));
    lon=-1;
}
void edgemake(int from,int to,int c)
{
    e[++lon].c=c;
    e[lon].to=to;
    e[lon].next=head[from];
    head[from]=lon;
}
void make(int from,int to,int c)
{
    edgemake(from,to,c);
    edgemake(to,from,0);
}

bool makelevel(int s,int t)
{
    memset(level,0,sizeof(level));
    int front=1,end=0;
    que[++end]=s;
    level[s]=1;
    while(front<=end)
    {
        int u=que[front++];
		if(u==t) return true;
        for(int k=head[u];k!=-1;k=e[k].next)
        {
            int v=e[k].to;
            if(!level[v]&&e[k].c)
            {
                que[++end]=v;
                level[v]=level[u]+1;
            }
        }
    }
	return false;
}

int dfs(int now,int t,int maxf)
{
    if(now==t||maxf==0) return maxf;
    int ret=0;
    for(int k=head[now];k!=-1;k=e[k].next)
    {
        int u=e[k].to;
		int f;
        if(level[u]==level[now]+1&&e[k].c)
        {
			f=dfs(u,t,min(e[k].c,maxf-ret));
            e[k].c-=f;
            e[k^1].c+=f;
            ret+=f;
            if(ret==maxf) return ret;
        }
    }
	if (ret==0) level[now] = 0;
    return ret;
}

int maxflow(int s,int t)
{
    int ret=0;
    while(makelevel(s,t))
    {
        ret+=dfs(s,t,inf);
    }
    return ret;
}
int main()
{
	int t;
	scanf("%d",&t);
	int sum=0;
	while(t--)
	{
		edgeini();
		int STA,END;
		int tot=0;
		sum++;
		int i,j;
		int d=0;
		int p,s,e;
		scanf("%d%d",&n,&m);
		for(i=1;i<=n;i++)
		{
			scanf("%d%d%d",&p,&s,&e);
			tot+=p;
			if(e>d) d=e;
			make(0,i,p);
			for(j=s;j<=e;j++) make(i,j+n,1);
		}
		STA=0;END=d+n+1;
		for(i=1;i<=d;i++) make(i+n,END,m);
		int ans=maxflow(STA,END);
		printf("Case %d: ",sum);
		if(ans==tot) printf("Yes
");
		else printf("No
");
		printf("
");
	}
	return 0;
}


 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=1050;
#define inf 10000000
int n,m;
int level[maxn],que[maxn*5];
int head[maxn],lon;
int a[maxn*2];
int min(int a,int b)
{
    if(a<b) return a;
    else return b;
}
struct
{
    int next,to,c;
}e[500100];
void edgeini()
{
    memset(head,-1,sizeof(head));
    lon=-1;
}
void edgemake(int from,int to,int c)
{
    e[++lon].c=c;
    e[lon].to=to;
    e[lon].next=head[from];
    head[from]=lon;
}
void make(int from,int to,int c)
{
    edgemake(from,to,c);
    edgemake(to,from,0);
}

bool makelevel(int s,int t)
{
    memset(level,0,sizeof(level));
    int front=1,end=0;
    que[++end]=s;
    level[s]=1;
    while(front<=end)
    {
        int u=que[front++];
        if(u==t) return true;
        for(int k=head[u];k!=-1;k=e[k].next)
        {
            int v=e[k].to;
            if(!level[v]&&e[k].c)
            {
                que[++end]=v;
                level[v]=level[u]+1;
            }
        }
    }
    return false;
}

int dfs(int now,int t,int maxf)
{
    if(now==t||maxf==0) return maxf;
    int ret=0;
    for(int k=head[now];k!=-1;k=e[k].next)
    {
        int u=e[k].to;
        int f;
        if(level[u]==level[now]+1&&e[k].c)
        {
            f=dfs(u,t,min(e[k].c,maxf-ret));
            e[k].c-=f;
            e[k^1].c+=f;
            ret+=f;
            if(ret==maxf) return ret;
        }
    }
    if (ret==0) level[now] = 0;
    return ret;
}

int maxflow(int s,int t)
{
    int ret=0;
    while(makelevel(s,t))
    {
        ret+=dfs(s,t,inf);
    }
    return ret;
}
int main()
{
    int t;
    while(scanf("%d%d",&n,&m)!=EOF)
    {	
        edgeini();
        int STA,END;
        int tot=0;
        int i,j;
		int si[maxn],ni,ei[maxn],ti;	
		int cnt=0;
		STA=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d%d%d%d",&si[i],&ni,&ei[i],&ti);
			make(STA,i,ni*ti);
            tot+=ni*ti;
			a[cnt++]=si[i];
			a[cnt++]=ei[i];
        }
		sort(a,a+cnt);
		j=1;
		for(i=1;i<cnt;i++)
			if(a[i]>a[i-1])
				a[j++]=a[i];
		cnt=j;
		END=n+cnt;		
		for(i=1;i<=cnt-1;i++)
		{
		   	make(n+i,END,m*(a[i]-a[i-1]));
			for(j=1;j<=n;j++)
				if(a[i-1]>=si[j]&&a[i]<=ei[j])
					make(j,n+i,inf);
		}
        int ans=maxflow(STA,END);
        if(ans==tot) printf("Yes
");
        else printf("No
");
    }
    return 0;
}


原文地址:https://www.cnblogs.com/vermouth/p/3710179.html