2013 Multi-University Training Contest 1 Vases and Flowers HDU 4614
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题意:给你N个花瓶,M次操作,每次输入K,如果K为1,那么再输入A,F,表示从A+1编号的花瓶开始,只要有空瓶,就插入一支花,如果F只花插完或者没有空瓶可以插了,就停止,输出最左边插入的和最右边插入的花瓶编号,当然,如果一支没插,或者F=0时,输出"Can not put any one.";如果K为2,再输入A,B,这时,将编号从A到B的所有花瓶中的花清空,输出清空的花的数量。
思路:明显的线段树。每个节点保存当前节点的左端点和右端点,以及flag标记,还有当前区间剩余的空瓶个数。当K为1时,先查询A+1~N-1之间的空瓶个数为tmp,如果tmp==0直接输出"Can not put any one.",否则在线段树中查找tmp空瓶的最左边和最右边的标号,分别为L,R,找到后更新区间[L,R]都为有花的情况;如果K为2,先查询当前A~B之间有多少花(等于区间花瓶总个数减去区间空瓶的个数),然后更新[A,B]为空瓶的情况。
#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define MAX 50005#define lson rt<<1#define rson rt<<1|1struct node{ int l,r,flag,sum,rest;} tree[MAX<<2];void pushUp(int rt){ tree[rt].rest=tree[lson].rest+tree[rson].rest; tree[rt].sum=tree[lson].sum+tree[rson].sum;}void pushDown(int rt){ int ls=lson,rs=rson; if(tree[rt].flag==-1) { tree[ls].flag=tree[rs].flag=-1; tree[ls].rest=tree[ls].r-tree[ls].l+1; tree[rs].rest=tree[rs].r-tree[rs].l+1; tree[rt].flag=0; } else if(tree[rt].flag==1) { tree[ls].flag=tree[rs].flag=1; tree[ls].rest=tree[rs].rest=0; tree[rt].flag=0; }}void build(int rt,int l,int r){ tree[rt].l=l,tree[rt].r=r; tree[rt].flag=0; tree[rt].rest=r-l+1; if(l==r) return ; int mid=(l+r)>>1; build(lson,l,mid); build(rson,mid+1,r);}int query(int rt,int l,int r){ if(r<l) return 0; if(tree[rt].l==l&&tree[rt].r==r) return tree[rt].rest; pushDown(rt); int mid=(tree[rt].l+tree[rt].r)>>1; if(r<=mid) return query(lson,l,r); else if(l>mid) return query(rson,l,r); else return query(lson,l,mid)+query(rson,mid+1,r);}int queryPos(int rt,int tag){ if(tree[rt].l==tree[rt].r) return tree[rt].l; pushDown(rt); int res=-1; if(tag<=tree[lson].rest) res=queryPos(lson,tag); else res=queryPos(rson,tag-tree[lson].rest); return res;}void update(int rt,int l,int r,int val){ if(tree[rt].flag==val) return ; if(tree[rt].l==l&&tree[rt].r==r) { if(val==1) { tree[rt].flag=1; tree[rt].rest=0; } else { tree[rt].flag=-1; tree[rt].rest=r-l+1; } return ; } pushDown(rt); int mid=(tree[rt].l+tree[rt].r)>>1; if(r<=mid) update(lson,l,r,val); else if(l>mid) update(rson,l,r,val); else { update(lson,l,mid,val); update(rson,mid+1,r,val); } pushUp(rt);}int main(){ int n,m,k,l,r; int T; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); build(1,1,n); while(m--) { scanf("%d%d%d",&k,&l,&r); if(k==1) { if(r==0) { puts("Can not put any one."); continue; } int tmp=query(1,1,l); if(tree[1].rest-tmp==0) { puts("Can not put any one."); continue; } if(tree[1].rest-tmp<r) r=tree[1].rest-tmp; int R=queryPos(1,tmp+r); int L=queryPos(1,tmp+1); update(1,L,R,1); printf("%d %d\n",L-1,R-1); } else { printf("%d\n",r-l+1-query(1,l+1,r+1)); update(1,l+1,r+1,-1); } } puts(""); } return 0;}
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