HDU 3998 Sequence 最长上升子序列+最大流

/*
很典型的题了，求最长上升子序列就不赘述了
现在要得到不重复利用元素，可以构造几个这样长度的序列
和同类型的求几条最短路类似，
Maxlen[i]表示到i的最长序列长度
Maxlen[i]=1的连源点
Maxlen[i]=maxlen的连汇点（maxlen为最长序列的长度）
对Maxlen[i]=Maxlen[j]+1(1<=j<i)连一条j到i的边
所有边边权为1,得到的最大流就是可以构造的最大个数
*/
#include <cstdio>
#include <iostream>
#include <memory.h>
#include<queue>
#include<set>
#include<ctime>
#include<algorithm>
#include<cmath>
#include<vector>
#define  LL __int64
using namespace std;
const int maxn=10009;
const LL inf=(1LL)<<35;
int aMaxlen[1009];
int b[1009];

struct edge
{
int v, next;
LL val;
} net[ 500010 ];
int n,m;
int level[maxn], Qu[maxn], out[maxn],next[maxn];
class Dinic {
public:
int end;
Dinic() {
end = 0;
memset( next, -1, sizeof(next) );
}
inline void insert( int x, int y, LL c) {
net[end].v = y, net[end].val = c,
net[end].next = next[x],
next[x] = end ++;
net[end].v = x, net[end].val = 0,
net[end].next = next[y],
next[y] = end ++;
}
bool BFS( int S, int E ) {
memset( level, -1, sizeof(level) );
int low = 0, high = 1;
Qu[0] = S, level[S] = 0;
for( ; low < high; ) {
int x = Qu[low];
for( int i = next[x]; i != -1; i = net[i].next ) {
if( net[i].val == 0 ) continue;
int y = net[i].v;
if( level[y] == -1 ) {
level[y] = level[x] + 1;
Qu[ high ++] = y;
}
}
low ++;
}
return level[E] != -1;
}

LL MaxFlow( int S, int E ){
LL maxflow = 0;
for( ; BFS(S, E) ; ) {
memcpy( out, next, sizeof(out) );
int now = -1;
for( ;; ) {
if( now < 0 ) {
int cur = out[S];
for(; cur != -1 ; cur = net[cur].next )
if( net[cur].val && out[net[cur].v] != -1 && level[net[cur].v] == 1 )
break;
if( cur >= 0 ) Qu[ ++now ] = cur, out[S] = net[cur].next;
else break;
}
int u = net[ Qu[now] ].v;
if( u == E ) {
LL flow = inf;
int index = -1;
for( int i = 0; i <= now; i ++ ) {
if( flow > net[ Qu[i] ].val )
flow = net[ Qu[i] ].val, index = i;
}
maxflow += flow;
for( int i = 0; i <= now; i ++ )
net[Qu[i]].val -= flow, net[Qu[i]^1].val += flow;
for( int i = 0; i <= now; i ++ ) {
if( net[ Qu[i] ].val == 0 ) {
now = index - 1;
break;
}
}
}
else{
int cur = out[u];
for(; cur != -1; cur = net[cur].next )
if (net[cur].val && out[net[cur].v] != -1 && level[u] + 1 == level[net[cur].v])
break;
if( cur != -1 )
Qu[++ now] = cur, out[u] = net[cur].next;
else out[u] = -1, now --;
}
}
}
return maxflow;
}
};
int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%d",&b[i]);
aMaxlen[1]=1;
for(int i=2;i<=n;i++)
{
int nTmp=0;
for(int j=1;j<i;j++)
if(b[i]>b[j])
{
if(aMaxlen[j]>nTmp)
nTmp=aMaxlen[j];
}
aMaxlen[i]=nTmp+1;
}
int nMax=-1;
for(int i=1;i<=n;i++)
if(nMax<aMaxlen[i])
nMax=aMaxlen[i];

Dinic my;
int end=1009;
for(int i=1;i<=n;i++)
{
if(aMaxlen[i]==nMax)
my.insert(i,end,1);
if(aMaxlen[i]==1)
my.insert(0,i,1);
}
for(int i=1;i<=n;i++)
for(int j=i+1;j<=n;j++)
{
if(aMaxlen[j]==aMaxlen[i]+1)
my.insert(i,j,1);
}
printf("%d\n%I64d\n",nMax,my.MaxFlow(0,end));
}
return 0;
}