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【loj】#6005. 「网络流 24 题」最长递增子序列(dp+最大流)

2017-07-19 16:49 316 查看
记录一个菜逼的成长。。

题目链接

网络流最重要的就是建边。

转化成二分图建边。

#include <bits/stdc++.h>
using namespace std;
#define ALL(v) (v).begin(),(v).end()
#define cl(a,b) memset(a,b,sizeof(a))
#define clr clear()
#define pb push_back
#define mp make_pair
#define fi first
#define se second
typedef long long LL;
typedef pair<int,int> PII;
const int INF = 0x3f3f3f3f;
const int MAX_V = 1000 + 10;
struct edge{
int to,cap,rev,flow;
edge(){}
edge(int _to,int _cap,int _rev,int _flow):to(_to),cap(_cap),rev(_rev),flow(_flow){}
};
vector<edge>G[MAX_V];
int level[MAX_V];
int iter[MAX_V];
void add(int from,int to,int cap,int flow = 0)
{
G[from].push_back(edge(to,cap,G[to].size(),0));
G[to].push_back(edge(from,0,G[from].size()-1,0));
}
void bfs(int s)
{
memset(level,-1,sizeof(level));
queue<int>que;
level[s] = 0;
que.push(s);
while(!que.empty()){
int f = que.front();
que.pop();
for( int i = 0; i < G[f].size(); i++ ){
edge &e = G[f][i];
if(e.cap > 0 && level[e.to] == -1){
level[e.to] = level[f] + 1;
que.push(e.to);
}
}
}
}
int dfs(int v,int t,int f)
{
if(v == t)return f;
for( int &i = iter[v]; i < G[v].size(); i++ ){
edge &e = G[v][i];
if(e.cap > 0 && level[v] < level[e.to]){
int d = dfs(e.to,t,min(e.cap,f));
if(d > 0){
e.cap -= d;e.flow += d;
G[e.to][e.rev].cap += d;
return d;
}
}
}
return 0;
}
int max_flow(int s,int t)
{
int flow = 0;
for(;;){
bfs(s);
if(level[t] == -1)return flow;
memset(iter,0,sizeof(iter));
int f;
while((f = dfs(s,t,INF)) > 0)
flow += f;
}
}
int s,t,n;
int a[MAX_V],dp[MAX_V],f[MAX_V];
int LIS(int *a,int n,int *dp)
{
int ret = 0;
for( int i = 1; i <= n; i++ )dp[i] = INF;
for( int i = 1; i <= n; i++ ){
int p = upper_bound(dp+1,dp+1+n,a[i]) - dp;
dp[p] = a[i];
f[i] = p;
ret = max(ret,p);
}
return ret;
}
void solve1(int lis)
{
for( int i = 1; i <= n; i++ ){
if(f[i] == 1)add(s,i,1);
add(i,i+n,1);
if(f[i] == lis)add(i+n,t,1);
for( int j = 1; j < i; j++ ){
if(a[j] <= a[i] && f[j] + 1 == f[i])add(j+n,i,1);
}
}
printf("%d\n",max_flow(s,t));
}
void solve2(int lis)
{
for( int i = s; i <= t; i++ )G[i].clr;
for( int i = 1; i <= n; i++ ){
if(i == 1 || i == n){
if(f[i] == 1)add(s,i,INF);
add(i,i+n,INF);
if(f[i] == lis)add(i+n,t,INF);
}
else {
if(f[i] == 1)add(s,i,1);
add(i,i+n,1);
if(f[i] == lis)add(i+n,t,1);
}
for( int j = 1; j < i; j++ )
if(a[j] <= a[i] && f[j] + 1 == f[i])add(j+n,i,1);
}
printf("%d\n",max_flow(s,t));
}
int main()
{
scanf("%d",&n);
s = 0,t = n + n + 1;
for( int i = 1; i <= n; i++ ){
scanf("%d",a+i);
}
int lis = LIS(a,n,dp);
cout<<lis<<endl;
if(lis == 1){
printf("%d\n%d\n",n,n);
return 0;
}
solve1(lis);
solve2(lis);
return 0;
}
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标签:  网络流 loj dp