hdu 3998 (dp+最大流)

题意：求最长上升子序列的长度和数量。

分析：用dp求出最长上升子序列m，dp数组存的就是该元素为子序列结尾的长度，源点与长度为1的点建边，长度为m的与汇点连边，然后枚举任意两个元素，ai,aj(ai>aj&&i>j&&dp[i]==dp[j]+1),j跟i连边，因为每个点只能选一次，所以边的容量都为1，求出最大流。

#include<stdio.h>
#include<string.h>
const int N=500;
const int inf=0x3fffffff;
int dis[N],gap[N],start,end,ans,dp[N],head[N],num;
struct edge
{
	int st,ed,flow,next;
}e[N*5];
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++;
	e[num].st=y;e[num].ed=x;e[num].flow=0;e[num].next=head[y];head[y]=num++;
}
int dfs(int u,int minflow)
{
    if(u==end)return minflow;
    int i,v,f,min_dis=ans-1,flow=0;
    for(i=head[u];i!=-1;i=e[i].next)
    {
        v=e[i].ed;
        if(e[i].flow<=0)continue;
        if(dis[v]+1==dis[u])
        {
            f=dfs(v,e[i].flow>minflow-flow?minflow-flow:e[i].flow);
            e[i].flow-=f;
            e[i^1].flow+=f;
            flow+=f;
            if(flow==minflow)break;
            if(dis[start]>=ans)return flow;
        }
        min_dis=min_dis>dis[v]?dis[v]:min_dis;
    }
    if(flow==0)
    {
        if(--gap[dis[u]]==0)
            dis[start]=ans;
        dis[u]=min_dis+1;
        gap[dis[u]]++;
    }
    return flow;
}
int isap()
{
    int maxflow=0;
    memset(gap,0,sizeof(gap));
    memset(dis,0,sizeof(dis));
    gap[0]=ans;
    while(dis[start]<ans)
        maxflow+=dfs(start,inf);
    return maxflow;
}
int main()
{
	int i,j,n,a[N],cot;
	while(scanf("%d",&n)!=-1)
	{
		cot=0;start=0;end=n+1;ans=end+1;num=0;
		memset(head,-1,sizeof(head));
		for(i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			dp[i]=1;
			for(j=1;j<i;j++)
			{if(a[i]>a[j]&&dp[i]<=dp[j])	dp[i]=dp[j]+1;}
			if(cot<dp[i])cot=dp[i];
		}
		printf("%d\n",cot);
		for(i=1;i<=n;i++)
		{
			if(dp[i]==1)
				addedge(start,i,1);
			if(dp[i]==cot)
				addedge(i,end,1);
			addedge(i,i,1);
		}
		for(i=1;i<=n;i++)
		{
			for(j=1;j<i;j++)
				if(a[j]<a[i]&&dp[j]+1==dp[i])
					addedge(j,i,1);
		}
		printf("%d\n",isap());
	}
	return 0;
}