uva1471 二叉搜索树

来源：互联网发布：2016淘宝女装销量前十编辑：程序博客网时间：2024/06/06 13:15

此题紫书上面有详细分析，关键是运用Set优化实现O(nlgn)复杂度

AC代码：

#include<cstdio>#include<set>#include<algorithm>using namespace std;const int maxn = 2e5+5;int num[maxn], h[maxn], g[maxn];//g[i] - num[i] is the Last//h[i] - num[i] is the Firststruct node{    int val, lenth;    node(){}    node(int val, int lenth):val(val), lenth(lenth){}    bool operator < (const node &p) const {        return val < p.val;    }};#define It set<node>::iteratorint solve(int n){    set<node> Set;    int ans = g[0];    Set.insert(node(num[0], g[0]));    for(int i = 1; i < n; ++i){        node c(num[i], g[i]);        It it = Set.lower_bound(c); //得到迭代器        bool ok = 1; //keep        if(it != Set.begin()){            node pre = *(--it);            ans = max(ans, pre.lenth + h[i]);            if(pre.lenth >= c.lenth) ok = 0;        }        if(ok){ //intsert C            Set.erase(c);            Set.insert(c);            it = Set.find(c);            it++;            while(it != Set.end() && it->lenth <= c.lenth) Set.erase(it++);        }    }    return ans;}int main(){    int T;    scanf("%d", &T);    while(T--){        int n;        scanf("%d", &n);        for(int i = 0; i < n; ++i) scanf("%d", &num[i]);        // 处理g[i]        g[0] = 1;        for(int i = 1; i < n; ++i) {            if(num[i] > num[i-1]) g[i] = g[i-1] + 1;            else g[i] = 1;        }        // 处理h[i]        h[n-1]=1;        for(int i = n-2; i >= 0; --i) {            if(num[i] < num[i+1]) h[i] = h[i+1] + 1;            else h[i] = 1;        }        printf("%d\n",solve(n));    }    return 0;}

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