一道线段树。lazy标记+位运算……(第一次写这个什么lazy标记,抄了一发题解)
我们发现:“或”操作在这里用正合适。
原题请戳// by Sirius_Ren#include#include #define N 100010using namespace std;int l,t,o,xx,yy,zz;char jy;struct segtree{ int left,right,num,lazy;}tree[N*4];void build(int l,int r,int pos){ tree[pos].left=l,tree[pos].right=r,tree[pos].num=2,tree[pos].lazy=false; if(l==r)return; int mid=(l+r)/2; build(l,mid,pos*2),build(mid+1,r,pos*2+1);}void push_down(int pos){ tree[pos*2].lazy=tree[pos*2+1].lazy=true; tree[pos*2+1].num=tree[pos*2].num=tree[pos].num; tree[pos].lazy=false;}void add(int pos){ if(tree[pos].left>=xx&&tree[pos].right<=yy){ tree[pos].lazy=true; tree[pos].num=(1< mid)add(pos*2+1); tree[pos].num=tree[pos*2].num|tree[pos*2+1].num;}int query(int pos){ if(tree[pos].left>=xx&&tree[pos].right<=yy)return tree[pos].num; if(tree[pos].lazy)push_down(pos); int mid=(tree[pos].left+tree[pos].right)/2; if(xx>mid)return query(pos*2+1); else if(yy<=mid)return query(pos*2); else return query(pos*2+1)|query(pos*2); }int split(){ int sum=query(1),ans=0; while(sum/=2)if(sum%2)ans++; return ans;}int main(){ scanf("%d%d%d",&l,&t,&o); build(1,l,1); for(int i=1;i<=o;i++){ scanf("\n%c%d%d",&jy,&xx,&yy); if(xx>yy)swap(xx,yy); if(jy=='C')scanf("%d",&zz),add(1); else printf("%d\n",split()); }}
晚上又写了一遍
很可惜,只刷到了Code Length第三… 省了一步建树。#include#include #include using namespace std;int L,o,xx=1,yy,zz=1,t[400001];char jy;bitset<400001>B;void pd(int p){t[p*2]=t[p*2+1]=t[p];B[p*2]=B[p*2+1]=true;B[p]=false;}void add(int l,int r,int p){ if(l>=xx&&r<=yy){B[p]=1;t[p]=1< m)add(m+1,r,p*2+1); t[p]=t[p*2]|t[p*2+1];}int q(int l,int r,int p){ if(xx<=l&&yy>=r)return t[p]; if(B[p])pd(p); int m=(l+r)/2; if(m =yy)return q(l,m,p*2); else return q(l,m,p*2)|q(m+1,r,p*2+1);}int main(){ scanf("%d%d%d",&L,&jy,&o); yy=L,add(1,L,1); while(o--){ scanf("\n%c%d%d",&jy,&xx,&yy); if(xx>yy)swap(xx,yy); if(jy=='C')scanf("%d",&zz),add(1,L,1); else{ int s=q(1,L,1),a=0; while(s){ if(s%2)a++;s/=2;} printf("%d\n",a); } }}