1741: 最长递增子序列问题——dp+网络流

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。 编程任务： 设计有效算法完成（1）（2）（3）提出的计算任务。

思路：

dp【i】表示以i为开头的LIS长度，用ans代表整个序列的LIS长度，对于第二问首先拆点，讲一个点拆成入点和出点，然后在入点和出点间连一条容量为1的有向边，之后在s与dp【i】为ans的入点之间连容量为1的有向边，然后在a【i】<a【j】且dp【i】=dp【j】+1的出点i、入点j之间连一条容量为1的有向边，最后在dp【i】=1的出点和t之间连一条容量为1的有向边。建完图后跑最大流即可。第三问在第二问的基础上讲1号和n号出入点之间的边的容量设置为INF，将s与1号店，n号店与t之间的边的容量设置为INF，跑最大流即可

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>

using namespace std;

const int maxn = 1000;
const int INF = 0x3f3f3f3f;

struct Edge {
    int from, to, cap, flow;
};

struct Dinic {
    int s, t;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];

    void init() {
        edges.clear();
        for (int i = 0; i < maxn; i++) G[i].clear();
    }

    void addedge(int from, int to, int cap) {
        edges.push_back(Edge{from, to, cap, 0});
        edges.push_back(Edge{to, from, 0, 0});
        int x = edges.size();
        G[from].push_back(x-2);
        G[to].push_back(x-1);
    }

    bool bfs() {
        memset(vis, 0, sizeof(vis));
        queue<int> q;
        q.push(s);
        d[s] = 0;
        vis[s] = 1;
        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 (!vis[e.to] && e.cap > e.flow) {
                    vis[e.to] = 1;
                    d[e.to] = d[x] + 1;
                    q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    int dfs(int x, int a) {
        if (x == t || a == 0) return a;
        int flow = 0, f;
        for (int &i = cur[x]; i < G[x].size(); i++) {
            Edge &e = edges[G[x][i]];
            if (d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) {
                e.flow += f;
                edges[G[x][i]^1].flow -= f;
                flow += f;
                a -= f;
                if (a == 0) break;
            }
        }
        return flow;
    }

    int maxflow(int s, int t) {
        this->s = s, this->t = t;
        int flow = 0;
        while (bfs()) {
            memset(cur, 0, sizeof(cur));
            flow += dfs(s, INF);
        }
        return flow;
    }
}ac, ac2;

int n, a[maxn], dp[maxn];

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= n; i++) dp[i] = 1;
    int ans = 1;
    int s = 0, t = 2 * n + 1;
    for (int i = n; i >= 1; i--) {
        for (int j = n; j > i; j--) {
            if (a[i] < a[j]) dp[i] = max(dp[i], dp[j] + 1);
        }
        ans = max(ans, dp[i]);
    }
    printf("%d\n", ans);
    if (ans == 1) {
        printf("%d\n%d\n", n, n);
        return 0;
    }
    ac.init();
    for (int i = 1; i <= n; i++) ac.addedge(i, i+n, 1);
    for (int i = 1; i <= n; i++) {
        for (int j = i + 1; j <= n; j++) {
            if (a[i] < a[j] && dp[i] == dp[j] + 1) {
                ac.addedge(i+n, j, 1);
            }
        }
    }
    for (int i = 1; i <= n; i++) {
        if (dp[i] == ans) ac.addedge(s, i, 1);
    }
    for (int i = 1; i <= n; i++) {
        if (dp[i] == 1) ac.addedge(i+n, t, 1);
    }
    printf("%d\n", ac.maxflow(s, t));
    ac2.init();
    ac2.addedge(1, 1+n, INF);
    for (int i = 2; i <= n-1; i++) ac2.addedge(i, i+n, 1);
    ac2.addedge(n, n+n, INF);
    for (int i = 1; i <= n; i++) {
        for (int j = i + 1; j <= n; j++) {
            if (a[i] < a[j] && dp[i] == dp[j] + 1) {
                ac2.addedge(i+n, j, 1);
            }
        }
    }
    for (int i = 1; i <= n; i++) {
        if (dp[i] == ans) {
            if (i == 1) ac2.addedge(s, 1, INF);
            else ac2.addedge(s, i, 1);
        }
    }
    for (int i = 1; i <= n; i++) {
        if (dp[i] == 1) {
            if (i == n) ac2.addedge(n+n, t, INF);
            else ac2.addedge(i+n, t, 1);
        }
    }
    printf("%d\n", ac2.maxflow(s, t));
    return 0;
}