hdu 4553 约会安排 (两个线段树维护区间)

• 题意: 一维区间,有申请一段连续区间的操作,优先级高的申请可以覆盖优先级低的,还可以清空一段区间的申请.问对每次申请的结果.
• 思路: 为两种申请都创建一颗线段树,每个节点保存左区间最长的,右区间最长的和总区间最长的连续区间

#include<string>
#include<vector>
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#define ll long long
#define FOR(i,l,r) for(int i = l ; i <= r ;++i )
#define inf 1<<30
#define EPS (1e-9)
#define lson(p) (p<<1)
#define rson(p) (p<<1|1)
using namespace std;
typedef pair<int,int> pii;
const int N = 1e5+30;
int n,m;
struct node{
    int l,r;
    int lsum,rsum,ssum;
}a[N<<2],b[N<<2];

void push_up(int rt,int l,int r){    
    int mid = (l+r)>>1;
    a[rt].lsum = a[lson(rt)].lsum;  // 左子树的左连续空间更新父亲
    a[rt].rsum = a[rson(rt)].rsum;  // 右子树的右连续空间更新父亲
    // 总连续空间为 max(左子树总连续,右子树总连续, 左子树右连续+右子树左连续)
    a[rt].ssum = max(max(a[lson(rt)].ssum,a[rson(rt)].ssum),a[lson(rt)].rsum+a[rson(rt)].lsum);
    if(a[lson(rt)].lsum == mid-l+1) a[rt].lsum += a[rson(rt)].lsum; // 左子树左连续包含了整个左子树,则父亲的左连续扩充右子树的左连续
    if(a[rson(rt)].rsum == r-mid)   a[rt].rsum += a[lson(rt)].rsum; 
    b[rt].lsum = b[lson(rt)].lsum;
    b[rt].rsum = b[rson(rt)].rsum;
    b[rt].ssum = max(max(b[lson(rt)].ssum,b[rson(rt)].ssum),b[lson(rt)].rsum+b[rson(rt)].lsum);
    if(b[lson(rt)].lsum == mid-l+1) b[rt].lsum += b[rson(rt)].lsum; 
    if(b[rson(rt)].rsum == r-mid)   b[rt].rsum += b[lson(rt)].rsum;

}

void push_down(int rt,int l,int r){
    int mid = (l+r)>>1;
    if(a[rt].ssum == r-l+1){ // 父亲被覆盖 ,子区间也被覆盖
        a[lson(rt)].lsum = a[lson(rt)].rsum = a[lson(rt)].ssum = mid-l+1;
        a[rson(rt)].lsum = a[rson(rt)].rsum = a[rson(rt)].ssum = r-mid;
    }else if(a[rt].ssum == 0){  // 父亲被情空,子区间清空
        a[lson(rt)].lsum = a[lson(rt)].rsum = a[lson(rt)].ssum = 0;
        a[rson(rt)].lsum = a[rson(rt)].rsum = a[rson(rt)].ssum = 0;      
    }
    if(b[rt].ssum == r-l+1){
        b[lson(rt)].lsum = b[lson(rt)].rsum = b[lson(rt)].ssum = mid-l+1;
        b[rson(rt)].lsum = b[rson(rt)].rsum = b[rson(rt)].ssum = r-mid;
    }else if(b[rt].ssum == 0){
        b[lson(rt)].lsum = b[lson(rt)].rsum = b[lson(rt)].ssum = 0;
        b[rson(rt)].lsum = b[rson(rt)].rsum = b[rson(rt)].ssum = 0;      
    }
}

void build(int r,int le,int re){
    a[r].lsum = b[r].lsum = (re-le+1);
    a[r].rsum = b[r].rsum = (re-le+1);
    a[r].ssum = b[r].ssum = (re-le+1);
    a[r].l = le,a[r].r = re;
    b[r].l = le,b[r].r = re;
    if(le==re)  return ;
    int mid = (le+re)>>1;
    build(lson(r),le,mid);
    build(rson(r),mid+1,re);
}


void update(int op,int l,int r,int rt){
    if(a[rt].l>=l && a[rt].r <=r){
        if(op==1){  // ds更新自己
            a[rt].lsum = a[rt].rsum = a[rt].ssum = 0;
        }else if(op==0){    // ns更新两棵树
            a[rt].lsum = a[rt].rsum = a[rt].ssum = 0;
            b[rt].lsum = b[rt].rsum = b[rt].ssum = 0;
        }else{  // 两棵树清空
            a[rt].lsum = a[rt].rsum = a[rt].ssum = a[rt].r-a[rt].l +1;
            b[rt].lsum = b[rt].rsum = b[rt].ssum = b[rt].r-b[rt].l +1;
        }
        return ;
    }
    int mid = (a[rt].l + a[rt].r)>>1;
    push_down(rt,a[rt].l,a[rt].r);  // 下方标记
    if(l<=mid)  update(op,l,r,lson(rt));
    if(r>mid)   update(op,l,r,rson(rt));
    push_up(rt,a[rt].l,a[rt].r);    // 上提标记
}

int query(int op,int rt,int len){
    if(a[rt].l==a[rt].r)    return a[rt].l;
    push_down(rt,a[rt].l,a[rt].r);
    int mid = (a[rt].l + a[rt].r) >> 1;
    if(op){     // ds 
        if(a[lson(rt)].ssum>=len)
            return query(op,lson(rt),len);
        else if(a[lson(rt)].rsum + a[rson(rt)].lsum >=len)  // 左子树的右区间加右子树的左区间满足
            return mid-a[lson(rt)].rsum +1; // 起点一定在左子树
        else
            return query(op,rson(rt),len);
    }else{      // ns
        if(b[lson(rt)].ssum>= len)
            return query(op,lson(rt),len);
        else if(b[lson(rt)].rsum + b[rson(rt)].lsum >=len)
            return mid-b[lson(rt)].rsum +1;
        else
            return query(op,rson(rt),len);
    }
}
char op[20];
void solve(int cas){
    printf("Case %d:
",cas);
    scanf("%d%d",&n,&m);
    build(1,1,n);
    int l,r;
    FOR(i,1,m){
        scanf("%s%d",op,&l);
        if(op[0]=='D'){
            if(a[1].ssum>=l){
                int pos = query(1,1,l);
                printf("%d,let's fly
",pos);
                update(1,pos,pos+l-1,1);
            }else{
                puts("fly with yourself");
            }
        }else if(op[0]=='N'){
            if(a[1].ssum>=l){
                int pos = query(1,1,l);
                printf("%d,don't put my gezi
",pos);
                update(0,pos,pos+l-1,1);
            }else if(b[1].ssum>=l){
                int pos = query(0,1,l);
                printf("%d,don't put my gezi
",pos);
                update(0,pos,pos+l-1,1);
            }else{
                puts("wait for me");
            }
        }else{
            scanf("%d",&r);
            update(2,l,r,1);
            printf("I am the hope of chinese chengxuyuan!!
");
        }
    }
}
int main(){
    int t;
    scanf("%d",&t);
    for(int i=1;i<=t;++i){
        solve(i);
    }
    return 0;
}