CodeForces

题意有点恶心，总结为： 每个点有一个值，这个值的定义是  这个点的度数（也就是有几个点跟他连着） *  从这个点往后找连着的编号递减的链长度

度数好算，输入时出现的次数就是度，，至于编号递减的链 在建边的时候建成有向边，从编号大的指向小的，

本来想跑dfs，后来发现从小的编号开始，直接一个递推过程就算完了

#include<iostream>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<cmath>#include<set>#include<queue>#include<stack>#include<map>using namespace std;typedef long long ll;const int maxn = 1e5 + 7, INF = 0x7f7f7f7f, mod = 1e9 + 7;int n, m;vector <int> vec[maxn];int g[maxn] = {0}, len[maxn];void init() {    scanf("%d %d", &n, &m);    for(int i = 0; i < m; ++i) {            int u, v;        scanf("%d %d", &u, &v);        g[u]++; g[v]++;        if(u > v) {            vec[u].push_back(v);        }        else {            vec[v].push_back(u);        }    }}int dfs(int id, int cnt) {    int ans = cnt;    for(int i = 0; i < vec[id].size(); ++i) {        ans = max(ans, len[vec[id][i]]);    }    return ans+1;}void solve() {    ll ans = 0LL;    memset(len, 0, sizeof len);    for(int i = 1; i <= n; ++i) {        len[i] = dfs(i, 0);        ans = max(ans, (ll)g[i] * (ll)len[i]);        //cout << " == " << g[i] << "  " << len[i] << " == " << endl;    }    cout << ans << endl;}int main() {    init();    solve();    return 0;}