[网络流24题]P2766 最长不下降子序列问题

网络流+LIS DP，一般般好题（重点在于看清题目）
https://www.luogu.com.cn/problem/P2766

题意

Tutorial

第一问用LIS求出

第二问首先计算 \(f(i)\)，从第 \(i\) 位开始的LIS最长多少。然后将每个点拆成入点和出点，限制每个点只能用一次。由于要求长度恰好为 \(s\) 的序列，所以如果第 \(i\) 个数满足 \(f(i) == s\) 则将 \(i\) 和 \(S\) 连边，如果第 \(i\) 个数满足 \(f(i) == 1\) 则将 \(i\) 和 \(T\) 连边。

第三问将 \(1\) 和 \(n\) 拆点间的边容量改为 \(INF\)，\(S\) 和 \(1\)， \(n\) 和 \(T\) 间的边容量改为 \(INF\) 即可

点击查看代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 500 + 5;
const int M = N * N;
const int INF = 0x3f3f3f3f;

int a[N];
int dp[N];
struct edge {
    int v, w, to;
} e[M * 2];
int pre[N<<1], cnt_edge, dep[N<<1];
int S, T, z, head[N<<1], sum;
int n, m, q[N<<1], cur[N<<1];
void add(int u, int v, int w) {
    // cout << u << " " << v << " " << cnt_edge << endl;
    e[cnt_edge] = {v, w, head[u]};
    head[u] = cnt_edge++;
    e[cnt_edge] = {u, 0, head[v]};
    head[v] = cnt_edge++;
}
bool bfs() {
    for (int i = 0; i <= T; i++) dep[i] = 0;
    dep[S] = 1;
    int l = 0, r = 1;
    q[r] = S;
    while (l < r) {
        int u = q[++l];
        for (int i = head[u]; i != -1; i = e[i].to) {
            int v = e[i].v;
            if (!dep[v] && e[i].w) dep[v] = dep[u] + 1, q[++r] = v;
        }
    }
    return dep[T];
}
int dfs(int u, int mi) {
    int res = 0;
    if (mi == 0 || u == T) return mi;
    for (int &i = cur[u]; i != -1; i = e[i].to) {
        int v = e[i].v;
        if (dep[u] + 1 == dep[v] && e[i].w) {
            int minn = dfs(v, min(mi - res, e[i].w));
            e[i].w -= minn;
            e[i ^ 1].w += minn;
            res += minn;
            if (res == mi) return res;
        }
    }
    if (res == 0) dep[u] = 0;
    return res;
}
int dinic() {
    ll res = 0;
    while (bfs()) {
        memcpy(cur, head, sizeof(head));
        //    cout<<res<<endl;
        res += dfs(S, INF);
    }
    return res;
}

int id(int x, int y) { return y*n+x; }
int main() {
    cin >> n;
    
    S = 2 * n + 1, T = S + 1;
    memset(dp, 0x3f, sizeof dp);
    memset(head, -1, sizeof head);
    for (int i = 1; i <= n; i++) cin >> a[i];

    if(n == 1){
        cout << "1\n1\n1";
        return 0;
    }
    for (int i = 1; i <= n; i++) {
        *upper_bound(dp + 1, dp + 1 + n, a[i]) = a[i];
    }

    int totlen = 0;
    for (int i = 1; i <= n && dp[i] != INF; i++) totlen++;
    cout << totlen << endl;

    for (int i = n; i; i--) {
        dp[i] = 1;
        for (int j = i + 1; j <= n; j++) {
            if (a[i] <= a[j]) {
                dp[i] = max(dp[i], dp[j] + 1);
            }
        }
    }

    // for (int i = 1; i <= n; i++) cout << dp[i] << " \n"[i == n];
    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 ) {
                add(id(i, 1), id(j, 0), 1);
            }
        }
    }
    vector<int> eid;
    for (int i = 1; i <= n; i++) {
        if (i == 1 || i == n) eid.push_back(cnt_edge);
        add(id(i, 0), id(i, 1), 1);
    }
    for (int i = 1; i <= n; i++) {
        if (dp[i] == totlen) {
            add(S, id(i, 0), 1);
            if (i == 1) eid.push_back(cnt_edge-2);
        }
        if (dp[i] == 1) {
            add(id(i, 1), T, 1);
            if (i == n) eid.push_back(cnt_edge-2);
        }
    }

    cout << dinic() << endl;
    for (int i = 0; i < cnt_edge; i += 2) {
        e[i].w += e[i ^ 1].w;
        e[i ^ 1].w = 0;
    }
    for(auto i:eid){
        // cout << i << " ";
        e[i].w = INF;
    }
    // cout << endl;
    // add(S, id(1, 0), INF);
    // add(id(n, 1), T, INF);
    // add(id(1, 1), id(1, 0), INF);
    // add(id(n, 1), id(n, 0), INF);
    cout << dinic() << endl;
}