Codeforces Round #338 (Div. 2) B. Longtail Hedgehog（LIS）

题意:

给定N≤105,M≤2×105的无向图,求图上节点标号LIS
输出答案maxni=1{LISi∗degi}

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 << endlconst 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;}