HDU4614 Vases and Flowers 二分+线段树

分析：感觉一看就是二分+线段树，没啥好想的，唯一注意，当开始摆花时，注意和最多能放的比大小

#include<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=5e4+5;
int n,m,T,c[N<<2],lz[N<<2];
void build(int rt,int l,int r){
  lz[rt]=-1;
  if(l==r){c[rt]=1;return;}
  int m=(l+r)>>1;
  build(rt<<1,l,m);
  build(rt<<1|1,m+1,r);
  c[rt]=c[rt<<1]+c[rt<<1|1];
}
void pushdown(int rt,int l,int r){
   if(lz[rt]!=-1){
     int m=(l+r)>>1;
     c[rt<<1]=lz[rt]*(m-l+1);
     c[rt<<1|1]=lz[rt]*(r-m);
     lz[rt<<1]=lz[rt<<1|1]=lz[rt];
     lz[rt]=-1;
   }
}
int t;
void modify(int rt,int l,int r,int x,int y){
  if(x<=l&&r<=y){
    c[rt]=t*(r-l+1);
    lz[rt]=t;
    return;
  }
  int m=(l+r)>>1;
  pushdown(rt,l,r);
  if(x<=m)modify(rt<<1,l,m,x,y);
  if(y>m)modify(rt<<1|1,m+1,r,x,y);
  c[rt]=c[rt<<1]+c[rt<<1|1];
}
int ask(int rt,int l,int r,int x,int y){
  if(x<=l&&r<=y)return c[rt];
  int m=(l+r)>>1,ans=0;
  pushdown(rt,l,r);
  if(x<=m)ans+=ask(rt<<1,l,m,x,y);
  if(y>m)ans+=ask(rt<<1|1,m+1,r,x,y);
  return ans;
}
int main(){
    scanf("%d",&T);
    while(T--){
      scanf("%d%d",&n,&m);
      build(1,1,n);
      while(m--){
         int op,x,y;
         scanf("%d%d%d",&op,&x,&y);
         if(op==2){
            ++x,++y;
            printf("%d\n",y-x+1-ask(1,1,n,x,y));
            t=1;modify(1,1,n,x,y);
            continue;
         }
         ++x;
         y=min(y,ask(1,1,n,x,n));
         if(!y){
          printf("Can not put any one.\n");
          continue;
         }
          int ans1,ans2;
          int l=x,r=n;
          while(l<r){
            int mid=(l+r)>>1,cur=ask(1,1,n,x,mid);
            if(!cur)l=mid+1;
            else r=mid;
          }
          ans1=(l+r)>>1;
          l=x,r=n;
          while(l<r){
            int mid=(l+r)>>1,cur=ask(1,1,n,x,mid);
            if(cur<y)l=mid+1;
            else r=mid;
          }
          ans2=(l+r)>>1;
          t=0;modify(1,1,n,ans1,ans2);
          printf("%d %d\n",--ans1,--ans2);
       }
       printf("\n");
    }
    return 0;
}