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  • bzoj 3251

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

    这道题在北京八十中的时候有人讲过。。 不过由于自己continue 写掉了一个所以调了很久。 

    做法是如果整个序列没有合法三角形的话,那么整个链长不超过50个(最大的情况是斐波那契数列) 所以大于50个一定成立, 小于50个排序扫一遍就好了

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    
    typedef long long ll;
    
    const ll maxn = 200010;
    const ll maxv = 50;
    
    ll int_get() {
        ll x = 0; char c = (char)getchar(); bool f = 0;
        while(!isdigit(c)) {
            if(c == '-') f = 1;
            c = (char)getchar();
        }
        while(isdigit(c)) {
            x = x * 10 + (ll)(c - '0');
            c = (char)getchar();
        }
        if(f) x = -x;
        return x;
    }
    
    struct edge {
        ll t; edge* next;
    }e[maxn * 2], *head[maxn]; ll ne = 0;
    
    void addedge(ll f, ll t) {
        e[ne].t = t, e[ne].next = head[f], head[f] = e + ne ++;
    }
    
    ll h[maxn]; 
    ll n, m, v[maxn], fa[maxn];
    
    void dfs(ll x, ll pre) {
        h[x] = h[pre] + 1;  fa[x] = pre;
        for(edge* p = head[x]; p; p = p-> next) {
            if(p-> t != pre) dfs(p-> t, x);
        }
    }
    
    void read() {
        n = int_get(); m = int_get();
        for(ll i = 1; i <= n; ++ i) v[i] = int_get();
        for(ll i = 1; i < n; ++ i) {
            ll f, t; 
            f = int_get(), t = int_get(); 
            addedge(f, t), addedge(t, f);
        }
        dfs(1, 0);
    }
    
    ll sta[maxn], top = 0;
    
    void sov() {
        while(m --) {
            ll opt = int_get(); 
            if(!opt) {
                ll a = int_get(), b = int_get(); 
                top = 0;
                if(h[a] < h[b]) swap(a, b);
                while(h[a] != h[b] && top <= maxv) sta[++ top] = v[a], a = fa[a];
                if(a != b) { 
                    while(a != b && top <= maxv) sta[++ top] = v[a], sta[++ top] = v[b], a = fa[a], b = fa[b];
                }
                sta[++ top] = v[a];
                if(top >= maxv) {
                    printf("Y
    "); continue;
                }
                sort(sta + 1, sta + top + 1); bool f = 0;
                for(ll i = 1; i < top - 1 && !f; ++ i) {
                    if(sta[i] + sta[i + 1] > sta[i + 2]) f = 1;
                }
                if(f) printf("Y
    ");
                else printf("N
    ");
            }
            else {
                ll p = int_get(), w = int_get();
                v[p] = w;
            }
        }
    }
    
    int main() {
        //freopen("test.in", "r", stdin);
        //freopen("test.out", "w", stdout);
        read(), sov();
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/ianaesthetic/p/4254718.html
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