BZOJ 3251: 树上三角形

最新推荐文章于 2019-11-23 09:52:54 发布

原创最新推荐文章于 2019-11-23 09:52:54 发布 · 200 阅读

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https://www.lydsy.com/JudgeOnline/problem.php?id=3251

考虑到一个序列如果不能组成三角形那么一定是fib序列
fib增长得很快并且值域只有1e9
那么50个一定有解
暴力跑路径大于50个直接跳出循环就可以了
需要注意的是判断时候要用上LL

#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <queue>
#include <cstdio>
#include <map>
#include <set>
#include <utility>
#include <stack>
#include <cstring>
#include <cmath>
#include <vector>
#include <ctime>
#include <bitset>
using namespace std;
#define pb push_back
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define ansn() printf("%d\n",ans)
#define lansn() printf("%lld\n",ans)
#define r0(i,n) for(int i=0;i<(n);++i)
#define r1(i,e) for(int i=1;i<=e;++i)
#define rn(i,e) for(int i=e;i>=1;--i)
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define lowbit(a) (a&(-a))
#define all(a) a.begin(),a.end()
#define pii pair<int,int>
#define pll pair<long long,long long>
#define mp(aa,bb) make_pair(aa,bb)
#define lrt rt<<1
#define rrt rt<<1|1
#define X first
#define Y second
#define PI (acos(-1.0))
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const ll mod = 1000000007 ;
const double eps=1e-9;
const int inf=0x3f3f3f3f;
//const ll infl = 100000000000000000;//1e17
const int maxn=  2e5+20;
const int maxm = 3e2+20;
//muv[i]=(p-(p/i))*muv[p%i]%p;
int in(int &ret) {
    char c;
    int sgn ;
    if(c=getchar(),c==EOF)return -1;
    while(c!='-'&&(c<'0'||c>'9'))c=getchar();
    sgn = (c=='-')?-1:1;
    ret = (c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');
    ret *=sgn;
    return 1;
}


int dep[maxn];
int val[maxn];
int fa[maxn];
vector<int>g[maxn];
void dfs(int u,int f) {
    int sz = g[u].size();
    r0(i,sz) {
        int v = g[u][i];
        if(v==f)continue;
        dep[v] = dep[u] + 1;
        fa[v] = u;
        dfs(v,u);
    }
}
vector<int>v;
bool check(ll a,ll b,ll c) {
    if(a+b>c&&a+c>b&&b+c>a)return 1;
    return 0;
}
void sol(int a,int b) {
    v.clear();
    int cnt = 0;
    while(a!=b&&cnt<=50) {
        if(dep[a]<=dep[b])swap(a,b);
        v.pb(val[a]);
        ++cnt;
        a = fa[a];
    }
    v.pb(val[b]);
    sort(all(v));
    int sz = v.size();
    for(int i=2; i<sz; ++i) {
        if(check(v[i-1],v[i-2],v[i])) {
            puts("Y");
            return ;
        }
    }
    puts("N");
}
int main() {
#ifdef LOCAL
    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
#endif // LOCAL

    int n,q;
    while(~sdd(n,q)) {
        r1(i,n)sd(val[i]);
        r0(i,n-1) {
            int u,v;
            sdd(u,v);
            g[u].pb(v);
//            g[v].pb(u);
        }
        dep[1] = 1;
        fa[1] = 0;
        dfs(1,0);
        while(q--) {
            int o;
            sd(o);
            int a,b;
            sdd(a,b);
            if(o&1)val[a] = b;
            else sol(a,b);
        }
    }
    return 0;
}