题意：
给定正整数序列 x1 ,... , xn 。
（1）计算其最长递增子序列的长度s。（严格递增）
（2）计算从给定的序列中最多可取出多少个长度为s的递增子序列。
（3）如果允许在取出的序列中多次使用x1和xn，则从给定序列中最多可取出多少个长度为s的递增子序列。
思路：
1. 可以抽象为求一张有向无环图的最多不相交路径，使得这些路径长度一样，且满足严格单调增的特性;
2. 构图比较巧妙：把每个点 i 拆分成 <i,a>, <i,b> 。假设最长上升子序列长度为 K. d[i] 表示从 i 开始到 n 最长上升子序列长度;
a. 由 s 向所有 d[i] = k 的 <i,a> 引弧，容量为 1，所有 d[i] = 1 的 <i, b> 向 t 引弧，容量为 1;
b. <i,a> 向 <i,b> 引弧，容量为 1. 当 i < j, 如果存在 d[i] == d[j] + 1 && seq[i] < seq[j] 时，由 <i,b> 向 <j,a> 引弧，容量为 1;
3. 求网络中的最大流即可，保证了每个点之被选择一次，并且经历了长度为 k 的路径。
4. 对于多次使用 x1, xn 的情况，把 s-><x1,a>, <x1,a>-><x1,b>, <xn,a>-><xn,b>, <xn,b>->t 修改成 INFS 求最大流即可;
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
const int MAXN = 1010;
const int INFS = 0x3FFFFFFF;
struct edge {
int from, to, cap, flow;
edge(int _from, int _to, int _cap, int _flow)
: from(_from), to(_to), cap(_cap), flow(_flow) {}
};
class Dinic {
public:
void initdata(int n, int s, int t) {
this->n = n, this->s = s, this->t = t;
edges.clear();
for (int i = 0; i < n; i++)
G[i].clear();
}
void addedge(int u, int v, int cap) {
edges.push_back(edge(u, v, cap, 0));
edges.push_back(edge(v, u, 0, 0));
G[u].push_back(edges.size() - 2);
G[v].push_back(edges.size() - 1);
}
bool BFS() {
for (int i = 0; i < n; i++)
vis[i] = false, d[i] = 0;
queue<int> Q;
Q.push(s);
vis[s] = true;
while (!Q.empty()) {
int x = Q.front(); Q.pop();
for (int i = 0; i < G[x].size(); i++) {
edge& e = edges[G[x][i]];
if (e.cap > e.flow && !vis[e.to]) {
vis[e.to] = true;
d[e.to] = d[x] + 1;
Q.push(e.to);
}
}
}
return vis[t];
}
int DFS(int x, int aug) {
if (x == t || aug == 0) return aug;
int flow = 0;
for (int i = 0; i < G[x].size(); i++) {
edge& e = edges[G[x][i]];
if (d[e.to] == d[x] + 1) {
int f = DFS(e.to, min(aug, e.cap - e.flow));
if (f <= 0) continue;
e.flow += f;
edges[G[x][i]^1].flow -= f;
flow += f;
aug -= f;
if (aug == 0) break;
}
}
return flow;
}
int maxflow() {
int flow = 0;
while (BFS()) {
flow += DFS(s, INFS);
}
return flow;
}
void change(int offset) {
for (int i = 0; i < edges.size(); i++) {
if (edges[i].from == s && edges[i].to == 1)
edges[i].cap = INFS;
if (edges[i].from == 1 && edges[i].to == offset + 1)
edges[i].cap = INFS;
if (edges[i].from == offset && edges[i].to == offset + offset)
edges[i].cap = INFS;
if (edges[i].from == offset + offset && edges[i].to == t)
edges[i].cap = INFS;
}
}
private:
vector<edge> edges;
vector<int> G[MAXN];
int n, s, t, d[MAXN];
bool vis[MAXN];
};
Dinic dc;
int dp[510], seq[510];
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++)
scanf("%d", &seq[i]);
memset(dp, 0, sizeof(dp));
dp[n] = 1;
int maxlen = 1;
for (int i = n-1; i >= 1; i--) {
dp[i] = 1;
for (int j = i+1; j <= n; j++)
if (seq[i] < seq[j])
dp[i] = max(dp[i], dp[j] + 1);
maxlen = max(maxlen, dp[i]);
}
int s = 0, t = 2*n + 1;
dc.initdata(t + 1, s, t);
for (int i = 1; i <= n; i++) {
dc.addedge(i, i + n, 1);
if (dp[i] == maxlen)
dc.addedge(s, i, 1);
if (dp[i] == 1)
dc.addedge(i + n, t, 1);
for (int j = i+1; j <= n; j++)
if (dp[i] == dp[j] + 1 && seq[i] < seq[j])
dc.addedge(i + n, j, 1);
}
int flow = dc.maxflow();
dc.change(n);
printf("%d\n%d\n%d\n", maxlen, flow, flow + dc.maxflow());
return 0;
}