Vases and Flowers HDU - 4614
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难点在第一问求第一朵和最后一朵花的位置 共有两种方法
一 二分求解
#include
using namespace std;
#define N 0x3f3f3f3f
struct node
{
int l;
int r;
int val;
int laz;
int pre;
};
node tree[200010];
int n,s,e;
void pushup(int cur)
{
tree[cur].val=tree[cur*2].val+tree[cur*2+1].val;
return;
}
void pushdown(int cur)
{
if(tree[cur].laz)
{
tree[cur*2].val=tree[cur].pre*(tree[cur*2].r-tree[cur*2].l+1);
tree[cur*2].laz=1;
tree[cur*2].pre=tree[cur].pre;
tree[cur*2+1].val=tree[cur].pre*(tree[cur*2+1].r-tree[cur*2+1].l+1);
tree[cur*2+1].laz=1;
tree[cur*2+1].pre=tree[cur].pre;
tree[cur].laz=0;
tree[cur].pre=0;
}
return;
}
void build(int l,int r,int cur)
{
int m;
tree[cur].l=l;
tree[cur].r=r;
tree[cur].val=0;
tree[cur].laz=0;
tree[cur].pre=0;
if(l==r) return;
m=(l+r)/2;
build(l,m,cur*2);
build(m+1,r,cur*2+1);
return;
}
void update(int val,int ll,int rr,int cur)
{
if(ll<=tree[cur].l&&tree[cur].r<=rr)
{
tree[cur].val=val*(tree[cur].r-tree[cur].l+1);
tree[cur].laz=1;
tree[cur].pre=val;
return;
}
pushdown(cur);
if(ll<=tree[cur*2].r) update(val,ll,rr,cur*2);
if(rr>=tree[cur*2+1].l) update(val,ll,rr,cur*2+1);
pushup(cur);
return;
}
int query(int ll,int rr,int cur)
{
int res;
if(ll<=tree[cur].l&&tree[cur].r<=rr)
{
return tree[cur].val;
}
pushdown(cur);
res=0;
if(ll<=tree[cur*2].r) res+=query(ll,rr,cur*2);
if(rr>=tree[cur*2+1].l) res+=query(ll,rr,cur*2+1);
return res;
}
int finds(int tar,int val)
{
int p,l,r,m,pre,ans;
p=tar,l=tar,r=n-1;
while(l<=r)
{
m=(l+r)/2;
pre=query(p,m,1);
if(m-p+1-pre==0)
{
l=m+1;
}
else if(m-p+1-pre==1)
{
ans=m;
r=m-1;
}
else
{
r=m-1;
}
}
return ans;
}
int finde(int tar,int val)
{
int p,l,r,m,pre,ans;
p=tar,l=tar,r=n-1;
while(l<=r)
{
m=(l+r)/2;
pre=query(p,m,1);
if(m-p+1-pre 二 乱搞
还是对区间的操作 在线段树遍历过程中每遇到一个”空区间“ 只要手里还有花就插 同时更新第一朵和最后一朵花的位置 插完就一路返回
#include
using namespace std;
#define N 0x3f3f3f3f
struct node
{
int l;
int r;
int val;
int laz;
int pre;
};
node tree[200010];
int n,s,e,val;
void pushup(int cur)
{
tree[cur].val=tree[cur*2].val+tree[cur*2+1].val;
return;
}
void pushdown(int cur)
{
if(tree[cur].laz)
{
tree[cur*2].val=tree[cur].pre*(tree[cur*2].r-tree[cur*2].l+1);
tree[cur*2].laz=1;
tree[cur*2].pre=tree[cur].pre;
tree[cur*2+1].val=tree[cur].pre*(tree[cur*2+1].r-tree[cur*2+1].l+1);
tree[cur*2+1].laz=1;
tree[cur*2+1].pre=tree[cur].pre;
tree[cur].laz=0;
tree[cur].pre=0;
}
return;
}
void build(int l,int r,int cur)
{
int m;
tree[cur].l=l;
tree[cur].r=r;
tree[cur].val=0;
tree[cur].laz=0;
tree[cur].pre=0;
if(l==r) return;
m=(l+r)/2;
build(l,m,cur*2);
build(m+1,r,cur*2+1);
return;
}
void update(int val,int ll,int rr,int cur)
{
if(ll<=tree[cur].l&&tree[cur].r<=rr)
{
tree[cur].val=val*(tree[cur].r-tree[cur].l+1);
tree[cur].laz=1;
tree[cur].pre=val;
return;
}
pushdown(cur);
if(ll<=tree[cur*2].r) update(val,ll,rr,cur*2);
if(rr>=tree[cur*2+1].l) update(val,ll,rr,cur*2+1);
pushup(cur);
return;
}
int queryI(int ll,int rr,int cur)
{
int res;
if(ll<=tree[cur].l&&tree[cur].r<=rr)
{
return tree[cur].val;
}
pushdown(cur);
res=0;
if(ll<=tree[cur*2].r) res+=queryI(ll,rr,cur*2);
if(rr>=tree[cur*2+1].l) res+=queryI(ll,rr,cur*2+1);
return res;
}
void queryII(int ll,int rr,int cur)
{
if(tree[cur].r-tree[cur].l+1==tree[cur].val||val<=0) return;
if(ll<=tree[cur].l&&tree[cur].r<=rr)
{
if(tree[cur].val==0)
{
s=min(s,tree[cur].l),e=max(e,min(tree[cur].l+val-1,tree[cur].r));
val-=(min(tree[cur].l+val-1,tree[cur].r)-tree[cur].l+1);
return;
}
}
if(tree[cur].l==tree[cur].r) return;
pushdown(cur);
if(ll<=tree[cur*2].r) queryII(ll,rr,cur*2);
if(rr>=tree[cur*2+1].l) queryII(ll,rr,cur*2+1);
return;
}
int main()
{
int t,q,op,tar,l,r;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&q);
build(0,n-1,1);
while(q--)
{
scanf("%d",&op);
if(op==1)
{
scanf("%d%d",&tar,&val);
s=N,e=-N;
queryII(tar,n-1,1);
if(s==N&&e==-N)
{
printf("Can not put any one.\n");
}
else
{
printf("%d %d\n",s,e);
update(1,s,e,1);
}
}
else
{
scanf("%d%d",&l,&r);
printf("%d\n",queryI(l,r,1));
update(0,l,r,1);
}
}
printf("\n");
}
return 0;
}