codeforces 615B

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题意：给出n个点是m条无向边，假设一条链长度是p，且这条链上的点是严格递增的，链的末端端点为u，求出最大的degree[u]*p

思路：dp[i]代表以i结尾的链的最大长度(不包括i本身)，那么我们对点输入的时候辩证一下顺序，然后对起点进行排序，最后答案就是(dp[i]+1)*degree[i]

#include<algorithm>#include<cstdio>#include<iostream>#include<cmath>#include<cstring>using namespace std;typedef long long ll;const int qq = 200000+10;int dp[qq/2];int degree[qq/2];struct Edge{int u,v;bool operator < (const Edge &a)const{if(a.u==u)return v<a.v;return u<a.u;}}edge[qq];int main(){int n,m;scanf("%d%d",&n,&m);memset(degree, 0, sizeof(degree));for(int i=0; i<m; ++i){scanf("%d%d",&edge[i].u,&edge[i].v);degree[edge[i].u]++,degree[edge[i].v]++;if(edge[i].u>edge[i].v)swap(edge[i].u, edge[i].v);}sort(edge, edge+m);memset(dp, 0, sizeof(dp));for(int i=0; i<m; ++i){int u=edge[i].u,v=edge[i].v;dp[v]=max(dp[v], dp[u]+1);}ll res=0;for(int i=1; i<=n; ++i)res=max(res, (ll)(dp[i]+1)*degree[i]);printf("%lld\n", res);return 0;}

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