HDU 1025-Constructing Roads In JGShining's Kingdom（最长不降子序列，线段树优化）

分析：

最长不降子序列，n很大o(n^2)肯定超，想到了小明序列那个题用线段树维护前面的最大值即可

该题也可用二分搜索来做。

注意问题输出时的坑，路复数后加s

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <string>
#include <cctype>
#include <complex>
#include <cassert>
#include <utility>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef pair<int,int> PII;
typedef long long ll;
#define lson l,m,rt<<1
#define pi acos(-1.0)
#define rson m+1,r,rt<<1|1
#define All 1,N,1
#define N 500010
#define read freopen("in.txt", "r", stdin)
const ll  INFll = 0x3f3f3f3f3f3f3f3fLL;
const int INF= 0x7ffffff;
const int mod =  ;
struct city{
  int p,r;
}t[N];
bool cmp(city x,city y){
    return x.p<y.p;
}
int maxv[N*],n,dp[N];
void pushup(int rt){
    maxv[rt]=max(maxv[rt<<],maxv[rt<<|]);
}
void build(int l,int r,int rt){
    maxv[rt]=;
    if(l==r){
        return;
    }
    int m=(l+r)>>;
    build(lson);
    build(rson);
}
void update(int p,int v,int l,int r,int rt){
      if(l==r){
        maxv[rt]=max(maxv[rt],v);
        return;
      }
      int m=(l+r)>>;
      if(p<=m)update(p,v,lson);
      else update(p,v,rson);
      pushup(rt);
}
int query(int L,int R,int l,int r,int rt){
    if(L<=l&&R>=r)
        return maxv[rt];
    int m=(l+r)>>;
    int ans=;
    if(L<=m)ans=max(ans,query(L,R,lson));
    if(R>m)ans=max(ans,query(L,R,rson));
    return ans;
}
int main()
{
    int ca=;
    while(~scanf("%d",&n)){
        memset(dp,,sizeof(dp));
        for(int i=;i<n;++i)
            scanf("%d%d",&t[i].p,&t[i].r);
        sort(t,t+n,cmp);
        build(,n,);
        int maxn=;
        for(int i=;i<n;++i){
            dp[i]=query(,t[i].r,,n,)+;
            maxn=max(maxn,dp[i]);
            update(t[i].r,dp[i],,n,);
        }
        printf("Case %d:\n",++ca);
        if(maxn>)
               printf("My king, at most %d roads can be built.\n",maxn);
        else
             printf("My king, at most %d road can be built.\n",maxn);
        printf("\n");
    }
return ;
}