Codeforces Round #338 (Div. 2) B. Longtail Hedgehog（LIS）-优快云博客

本文链接：https://blog.youkuaiyun.com/lwt36/article/details/50490016

本文介绍了一种算法，用于解决给定大规模无向图中寻找最长递增子序列(LIS)的问题，并计算每个节点的LIS与其度数的乘积的最大值。通过O(m)的时间复杂度实现高效的求解。

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题意:

$给定N\le 10^5, M \le 2×10^5的无向图, 求图上节点标号LIS$
$输出答案max_{i=1}^n\{LIS_i*deg_i\}$

分析:

$f[i]:=以节点i结尾的LIS长度, 直接转移即可, 总复杂度O(m)$
$扫一遍f[i]数组更新答案$

代码:

//
//  Created by TaoSama on 2016-01-08
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m;
vector<int> G[N];

int f[N];

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%d%d", &n, &m) == 2){
        for(int i = 1; i <= n; ++i) G[i].clear();
        for(int i = 1; i <= m; ++i){
            int u, v; scanf("%d%d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        for(int i = 1; i <= n; ++i) f[i] = 1;
        for(int u = 1; u <= n; ++u)
            for(int v : G[u])
                if(v > u) f[v] = max(f[v], f[u] + 1);

        long long ans = 0;
        for(int i = 1; i <= n; ++i)
            ans = max(ans, 1LL * f[i] * G[i].size());
        printf("%I64d\n", ans);
    }
    return 0;
}