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心若平似镜、何题不AC。

uestc 1425 Another LCIS 线段树 区间合并

结点记录的信息:

lics:区间lics的大小

lv、rv:区间左右端点的值

lpart:以左端点为起点的ics的大小

rpart:以右端点为终点的ics的大小

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define pb(a) push_back(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
#define clr0(a) memset(a,0,sizeof(a))
#define clr1(a) memset(a,-1,sizeof(a))
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("d:\\in.txt","r",stdin);
    freopen("d:\\out1.txt","w",stdout);
#endif
}
char getch() {
    char ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='\n')return ch;
    }
    return EOF;
}
struct node
{
    int lics,lpart,rpart,lv,rv,len;
    node(){}
    node(int lics,int lpart,int rpart,int lv,int rv,int len):lics(lics),lpart(lpart),rpart(rpart),lv(lv),rv(rv),len(len){}
};
const int maxn=100010;
node tree[maxn<<2];
int add[maxn<<2];
int Merge(node &idx,node &a,node &b)
{
    if(a.rv<b.lv)
    {
        idx.lics=max(a.lics,b.lics,a.rpart+b.lpart);
        idx.lpart=a.lpart==a.len?a.lpart+b.lpart:a.lpart;
        idx.rpart=b.rpart==b.len?b.rpart+a.rpart:b.rpart;
    }else
    {
        idx.lics=max(a.lics,b.lics);
        idx.lpart=a.lpart;
        idx.rpart=b.rpart;
    }
    idx.lv=a.lv;idx.rv=b.rv;
    idx.len=a.len+b.len;
}
int PushDown(int idx)
{
    if(add[idx]!=0)
    {
        add[idx<<1]+=add[idx];
        add[idx<<1|1]+=add[idx];
        tree[idx<<1].lv+=add[idx];
        tree[idx<<1].rv+=add[idx];
        tree[idx<<1|1].lv+=add[idx];
        tree[idx<<1|1].rv+=add[idx];
        add[idx]=0;
    }
    return 0;
}
int Build(int idx,int l,int r)
{
    if(l==r)
    {
        int v;
        scanf("%d",&v);
        tree[idx]=node(1,1,1,v,v,1);
        add[idx]=0;
        return 0;
    }
    int mid=(r+l)>>1;
    Build(lson);Build(rson);
    Merge(tree[idx],tree[idx<<1],tree[idx<<1|1]);
    add[idx]=0;
    return 0;
}
int Update(int idx,int l,int r,int tl,int tr,int v)
{
    if(tl<=l&&tr>=r)
    {
        add[idx]+=v;
        tree[idx].lv+=v;tree[idx].rv+=v;
        return 0;
    }
    PushDown(idx);
    int mid=(r+l)>>1;
    if(tl<=mid)Update(lson,tl,tr,v);
    if(tr>mid)Update(rson,tl,tr,v);
    Merge(tree[idx],tree[idx<<1],tree[idx<<1|1]);
    return 0;
}
node Quary(int idx,int l,int r,int tl,int tr)
{
    if(tl<=l&&tr>=r)
    {
        return tree[idx];
    }
    PushDown(idx);
    int mid=(r+l)>>1;
    node q;
    if(tl<=mid&&tr>mid)
    {
        node a,b;
        a=Quary(lson,tl,tr);
        b=Quary(rson,tl,tr);
        Merge(q,a,b);
    }else if(tl<=mid)
    {
        q= Quary(lson,tl,tr);
    }else q= Quary(rson,tl,tr);
    Merge(tree[idx],tree[idx<<1],tree[idx<<1|1]);
    return q;
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++)
    {
        int n,q;
        scanf("%d%d",&n,&q);
        Build(1,1,n);
        printf("Case #%d:\n",ca);
        for(int i=1;i<=q;i++)
        {
            char op[10];
            scanf("%s",op);
            if(op[0]=='a')
            {
                int l,r,v;
                scanf("%d%d%d",&l,&r,&v);
                Update(1,1,n,l,r,v);
            }else
            {
                int l,r;
                scanf("%d%d",&l,&r);
                node x=Quary(1,1,n,l,r);
                printf("%d\n",x.lics);
            }
        }
    }
    return 0;
}
View Code

 

posted on 2013-08-17 13:38  BMan、  阅读(224)  评论(0编辑  收藏  举报

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