kuangbin带你飞--线段树二
j
题意：
每一个人都有一个boss，没有boss的那一个人是最终大boss，每一次发放任务个x，他的所有的以他为boss的人也就是说子树（包括自己）全部都开始做这个任务。
我们可以看到这一棵树是无序的，而且是并不是二叉树，所以就需要转换。
既然跟子树有关，我们就深度优先遍历，用dfs来建立序列，一个点有开始和结束的编号，他的所有子树的编号一定会在开始和节点之间，这样子就可以做到区间更新。
所以我们根据dfs序列来进行建立线段树的操作，单点更新到开始的编号，区间更新的话就是这个节点开始到结束的编号。
这里还利用了类似并查集的原理来寻炸根节点。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<map>
//#include<regex>
#include<cstdio>
#define up(i,a,b) for(int i=a;i<b;i++)
#define dw(i,a,b) for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
int read()
{
char ch = getchar(); int x = 0, f = 1;
while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return x * f;
}
typedef pair<int, int> pir;
const int N = 50010;
int T;
int n, m,tot,cnt;
int st[N], ed[N];
struct {
int to, next;
}edge[N];
int head[N];
int tree[N<<2];
int lazy[N << 2];
int uset[N];
int find_set(int x)
{
if (uset[x] != x)
uset[x] = find_set(uset[x]);
return uset[x];
}
void build(int l, int r, int root)
{
lazy[root] = -1;
tree[root] = -1;
if (l == r)
{
return;
}
int mid = (l + r) >> 1;
build(l, mid, root << 1);
build(mid + 1, r, root << 1 | 1);
}
void pushdown(int root)
{
if (lazy[root] != -1)
{
tree[root << 1] = lazy[root];
tree[root << 1 | 1] = lazy[root];
lazy[root << 1] = lazy[root];
lazy[root << 1 | 1] = lazy[root];
lazy[root] = -1;
}
}
void addedge(int f,int t)
{
edge[tot].to = t;
edge[tot].next = head[f];
head[f] = tot++;
}
void dfs(int p)
{
st[p] = ++cnt;
int i = head[p];
while(i!=-1)
{
dfs(edge[i].to);
i = edge[i].next;
}
ed[p] = cnt;
}
void update(int l, int r, int root, int lf, int rt, int val)
{
if (lf <= l && r <= rt)
{
tree[root] = val; lazy[root] = val; return;
}
pushdown(root);
int mid = (l + r) >> 1;
if (lf <= mid)update(l, mid, root << 1, lf, rt, val);
if (rt > mid)update(mid + 1, r, root << 1 | 1, lf, rt, val);
}
void querry(int l, int r, int root ,int pos)
{
if (l == r)
{
printf("%d\n", tree[root]);
return;
}
pushdown(root);
int mid = (l + r) >> 1;
if (pos <= mid)return querry(l, mid, root << 1, pos);
else return querry(mid + 1, r, root << 1 | 1, pos);
}
int main()
{
scanf("%d", &T);
upd(p, 1, T)
{
scanf("%d", &n);
upd(i, 0, n)uset[i] = i,head[i]=-1;
tot = 0; cnt = 0;
int x, y;
// cout << tot << " " << cnt << endl;
up(i, 0, n - 1)
{
scanf("%d %d",&x,&y); addedge(y, x); uset[x] = y;
}
int rt = find_set(1);
dfs(rt);
build(1, cnt, 1);
scanf("%d", &m);
char op[3];
printf("Case #%d:\n", p);
while (m--)
{
scanf("%s", op);
if (op[0] == 'C')
{
scanf("%d", &x);
querry(1, n, 1, st[x]);
// printf("%d", querry(1, cnt, 1, st[x]));
}
else
{
scanf("%d %d", &x, &y);
update(1, cnt, 1, st[x], ed[x], y);
}
}
}
return 0;
}
K
题意
给出一系列操作，有给这一个区间乘以一个数字的，有给这一个区间加上一个数字的，有询问这一个区间，平方和，立方和的。
这里询问操作就不再赘述，网上题解写的好的很。
主要是细节 ，不要粗心。
先做乘法后做加法。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<map>
//#include<regex>
#include<cstdio>
#define up(i,a,b) for(int i=a;i<b;i++)
#define dw(i,a,b) for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
int read()
{
char ch = getchar(); int x = 0, f = 1;
while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
const int N = 1e5 + 10;
const int mod = 10007;
const int MOD = 10007;
ll sum1[N << 2];
ll sum2[N << 2];
ll sum3[N << 2];
ll st[N << 2];
ll mul[N << 2];
ll add[N << 2];
int x, y, z, op;
int n, m;
ll madd(ll x, ll y)
{
return (x%mod + y % mod) % mod;
}
ll mmul(ll x, ll y)
{
return x * y%mod;
}
void pushup(int root)
{
sum1[root] = madd(sum1[root << 1], sum1[root << 1 | 1]);
sum2[root] = madd(sum2[root << 1], sum2[root << 1 | 1]);
sum3[root] = madd(sum3[root << 1], sum3[root << 1 | 1]);
}
void pushdown(int rt, int len)
{
int root = rt;
if (st[root])
{
add[root << 1] = add[root << 1 | 1] = 0;
mul[root << 1] = mul[root << 1 | 1] = 1;
st[root << 1] = st[root];
st[root << 1 | 1] = st[root];
ll lenl = len - (len >> 1); ll lenr = (len >> 1);
sum1[root << 1] = mmul(st[root], lenl%mod);
sum1[root << 1 | 1] = mmul(st[root], lenr%mod);
sum2[root << 1] = lenl % mod * (st[root] * st[root] % mod) % mod;
sum2[root << 1 | 1] = lenr % mod * (st[root] * st[root] % mod) % mod;
sum3[root << 1] = lenl % mod * ((st[root] * st[root]) % mod*st[root] % mod) % mod;
sum3[root << 1 | 1] = lenr % mod * ((st[root] * st[root]) % mod*st[root] % mod) % mod;
st[root] = 0;
}
if (mul[rt] != 1) { //这个就是mul[rt] != 1 ， 当时我这里没注意所以TLE了
mul[rt << 1] = (mul[rt << 1] * mul[rt]) % MOD;
mul[rt << 1 | 1] = (mul[rt << 1 | 1] * mul[rt]) % MOD;
if (add[rt << 1]) //注意这个也要下放
add[rt << 1] = (add[rt << 1] * mul[rt]) % MOD;
if (add[rt << 1 | 1])
add[rt << 1 | 1] = (add[rt << 1 | 1] * mul[rt]) % MOD;
ll tmp = (((mul[rt] * mul[rt]) % MOD * mul[rt]) % MOD);
sum1[rt << 1] = (sum1[rt << 1] * mul[rt]) % MOD;
sum1[rt << 1 | 1] = (sum1[rt << 1 | 1] * mul[rt]) % MOD;
sum2[rt << 1] = (sum2[rt << 1] % MOD) * ((mul[rt] * mul[rt]) % MOD) % MOD;
sum2[rt << 1 | 1] = (sum2[rt << 1 | 1] % MOD) * ((mul[rt] * mul[rt]) % MOD) % MOD;
sum3[rt << 1] = (sum3[rt << 1] % MOD) * tmp % MOD;
sum3[rt << 1 | 1] = (sum3[rt << 1 | 1] % MOD) * tmp % MOD;
mul[rt] = 1;
}
if (add[root])
{
add[root << 1] = madd(add[root << 1], add[root]);
add[root << 1 | 1] = madd(add[root << 1 | 1], add[root]);
ll tp2 = add[root] * add[root] % mod;
ll tp3 = tp2 * add[root] % mod;
ll tmp = (add[rt] * add[rt] % MOD) * add[rt] % MOD; //注意sum3 , sum2 , sum1的先后顺序
sum3[rt << 1] = (sum3[rt << 1] + (tmp * (len - (len >> 1)) % MOD) + 3 * add[rt] * ((sum2[rt << 1] + sum1[rt << 1] * add[rt]) % MOD)) % MOD;
sum3[rt << 1 | 1] = (sum3[rt << 1 | 1] + (tmp * (len >> 1) % MOD) + 3 * add[rt] * ((sum2[rt << 1 | 1] + sum1[rt << 1 | 1] * add[rt]) % MOD)) % MOD;
sum2[root << 1] = (sum2[root << 1] + 2ll * sum1[root << 1] * add[root] % mod + (1ll * (len - (len >> 1))*tp2) % mod) % mod;
sum2[root << 1 | 1] = (sum2[root << 1 | 1] + 2ll * sum1[root << 1 | 1] * add[root] % mod + 1ll * (len >> 1)*tp2 % mod) % mod;
sum1[root << 1] = madd(sum1[root << 1], mmul(len - (len >> 1), add[root]));
sum1[root << 1 | 1] = madd(sum1[root << 1 | 1], mmul((len >> 1), add[root]));
add[root] = 0;
}
}
void build(int l, int r, int root)
{
st[root] = 0;
mul[root] = 1;
add[root] = 0;
sum1[root] = sum2[root] = sum3[root] = 0;
if (l == r)return;
int mid = (l + r) >> 1;
build(lson);
build(rson);
pushup(root);
}
void update(int l, int r, int root, int lf, int rt, int op, int val)
{
if (lf <= l && r <= rt)
{
if (op == 1)
{
add[root] = (add[root] + val) % mod;
ll tp2 = 1ll * val*val % mod;
ll tp3 = 1ll * tp2 * val % mod;
sum3[root] = (sum3[root] + 3ll * sum2[root] * val%mod + 3ll * sum1[root] * tp2%mod + tp3 * 1ll * (r - l + 1) % mod) % mod;
sum2[root] = (sum2[root] + 2ll * sum1[root] * val%mod + tp2 * 1ll * (r - l + 1) % mod) % mod;
sum1[root] = (sum1[root] + 1ll * (r - l + 1)*val) % mod;
}
if (op == 2)
{
mul[root] = (1ll * mul[root] * val) % mod;
add[root] = (1ll * add[root] * val) % mod;
sum3[root] = (sum3[root] * ((1ll * val*val%mod)*val%mod)) % mod;
sum2[root] = (sum2[root] * (1ll * val*val%mod)) % mod;
sum1[root] = (1ll * sum1[root] * val) % mod;
}
if (op == 3)
{
st[root] = val;
add[root] = 0;
mul[root] = 1;
sum3[root] = 1ll * (r - l + 1)*((1ll * val*val) % mod*val%mod) % mod;
sum2[root] = 1ll * (r - l + 1)*(1ll * val*val%mod) % mod;
sum1[root] = 1ll * (r - l + 1)*val%mod;
}
return;
}
pushdown(root, (r - l + 1));
int mid = (l + r) >> 1;
if (lf <= mid)update(lson, lf, rt, op, val);
if (rt > mid)update(rson, lf, rt, op, val);
pushup(root);
}
ll querry(int l, int r, int root, int lf, int rt, int p)
{
if (lf <= l && r <= rt)
{
if (p == 1)return sum1[root] % mod;
if (p == 2)return sum2[root] % mod;
if (p == 3)return sum3[root] % mod;
}
pushdown(root, r - l + 1);
int mid = (l + r) >> 1;
ll ans = 0;
if (lf <= mid)ans += querry(lson, lf, rt, p);
if (rt > mid)ans += querry(rson, lf, rt, p);
ans %= mod;
return ans;
}
int main()
{
while (cin >> n >> m && (n + m))
{
build(1, n, 1);
up(i, 0, m)
{
op = read(), x = read(), y = read(), z = read();
if (op == 4)
{
printf("%lld\n", querry(1, n, 1, x, y, z));
}
else
{
update(1, n, 1, x, y, op, z);
}
}
}
return 0;
}
L 题意
给出几个操作，第一个操作是把花插入花瓶，并指定位置，一直插到n-1
或者花不够了结束。如果遇到花瓶有花了，跳过。也就是从pos到n-1，找空位置插入。
询问就是区间内有几个空的花瓶。
我们反过来用1表示空，0表示有。
这样区间和就可以表示这个区间还有几个空的花瓶了。
我们这里运用二分的思想。对于一个区间来说，他的和是上升不会下降的。
例如11100011101，即便是被插了花，和来说也是不严格单调递增的。
所以我们能用二分。先二分起点，也就是querry查询sum=1，再二分终点，查询sum=花数目，最后更新这个区间，置零即可。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<map>
//#include<regex>
#include<cstdio>
#define up(i,a,b) for(int i=a;i<b;i++)
#define dw(i,a,b) for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
int read()
{
char ch = getchar(); int x = 0, f = 1;
while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return x * f;
}
typedef pair<int, int> pir;
const int N = 5e4 + 10;
int T, n, m,op;
int tree[N << 2];
int lazy[N << 2];
void pushup(int root)
{
tree[root] = tree[lrt] + tree[rrt];
}
void pushdown(int root,int len)
{
if (lazy[root]!=-1)
{
tree[lrt] = (len - (len >> 1))*lazy[root];
tree[rrt] = (len >> 1)*lazy[root];
lazy[lrt] = lazy[rrt] = lazy[root];
lazy[root] = -1;
}
}
void build(int l, int r, int root)
{
lazy[root] = -1;
if (l == r)
{
tree[root] = 1;
return;
}
int mid = (l + r) >> 1;
build(lson);
build(rson);
pushup(root);
}
void update(int l, int r, int root, int lf, int rt,int val)
{
if (lf <= l && r <= rt)
{
tree[root] = (r - l + 1)*val;
lazy[root] = val;
return;
}
pushdown(root, r - l + 1);
int mid = (l + r) >> 1;
if (lf <= mid)update(lson, lf, rt, val);
if (rt > mid)update(rson, lf, rt, val);
pushup(root);
}
int querry(int l, int r, int root, int lf, int rt)
{
if (lf <= l && r <= rt)
{
return tree[root];
}
pushdown(root, r - l + 1);
int mid = (l + r) >> 1;
int ans = 0;
if (lf <= mid)ans += querry(lson, lf, rt);
if (rt > mid)ans += querry(rson, lf, rt);
return ans;
}
int binarysearch(int a,int b,int num)
{
int templ = a-1; int tempr = b+1;
while (tempr -1 > templ)
{
int mid = (templ + tempr) >> 1;
if (querry(0, n - 1, 1, a, mid) < num)templ = mid;
else tempr = mid;
}
return tempr;
}
int main()
{
T = read();
while (T--)
{
int x, y;
n = read(), m = read();
build(0, n - 1, 1);
while (m--)
{
op = read(); x = read(), y = read();
if (op == 1)
{
int temp = querry(0, n - 1, 1, x, n - 1);
int ansl = binarysearch(x, n - 1, 1);
int ansr = binarysearch(x, n - 1, min(temp, y));
if (ansl == n)printf("Can not put any one.\n");
else {
printf("%d %d\n", ansl, ansr);
update(0, n - 1, 1, ansl, ansr, 0);
}
}
if (op == 2)
{
int sum = querry(0, n - 1, 1, x, y);
update(0, n - 1, 1, x, y, 1);
printf("%d\n", (y-x+1)-sum);
}
}
cout << endl;
}
return 0;
}
M
题意：屌丝有屌丝想要约的一段时间，自己安排某个区间刚好放下这一段时间，而对于女神来说，女神与女神之间互相不干扰，不能女生用另外一个女神的时间然而可以用屌丝已经安排好的时间，挤掉屌丝。针对q个询问，是否能安排出来时间。
我这里运用了两颗线段树，实际代码可能更短，分别维护屌丝树和女神树，其他的都是基本操作，代码略长而已。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<map>
//#include<regex>
#include<cstdio>
#define up(i,a,b) for(int i=a;i<b;i++)
#define dw(i,a,b) for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
int read()
{
char ch = getchar(); int x = 0, f = 1;
while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 1e5 + 10;
struct { int l, r,m; }dstree[N<<2],nstree[N<<2];
int dslazy[N << 2],nslazy[N<<2];
int CASE,t,n;
void dspushup(int root,int len)
{
dstree[root].l = dstree[root << 1].l;
dstree[root].r = dstree[root << 1 | 1].r;
dstree[root].m = max(dstree[lrt].m, dstree[rrt].m);
if (dstree[lrt].l == (len-(len>>1)))dstree[root].l += dstree[rrt].l;
if (dstree[rrt].r == (len>>1))dstree[root].r += dstree[lrt].r;
dstree[root].m = max(dstree[root].m, dstree[lrt].r + dstree[rrt].l);
}
void nspushup(int root,int len)
{
nstree[root].l = nstree[lrt].l;
nstree[root].r = nstree[rrt].r;
nstree[root].m = max(nstree[lrt].m, nstree[rrt].m);
if (nstree[lrt].l == (len - (len >> 1)))nstree[root].l += nstree[rrt].l;
if (nstree[rrt].r == (len >> 1))nstree[root].r += nstree[lrt].r;
nstree[root].m = max(nstree[root].m, nstree[lrt].r + nstree[rrt].l);
}
void dspushdown(int root, int len)
{
if (dslazy[root] == -1)return;
if (dslazy[root] ==1)
{
dstree[lrt].l = dstree[lrt].r=dstree[lrt].m=(len - (len >> 1));
dstree[rrt].r = dstree[rrt].l = dstree[rrt].m = (len >> 1);
dslazy[lrt] = dslazy[root];
dslazy[rrt] = dslazy[root];
dslazy[root] = -1;
}
else
{
dstree[lrt].l = dstree[lrt].r = dstree[lrt].m = 0;
dstree[rrt].l = dstree[rrt].r = dstree[rrt].m = 0;
dslazy[lrt] = dslazy[root];
dslazy[rrt] = dslazy[root];
dslazy[root] = -1;
}
}
void nspushdown(int root, int len)
{
if (nslazy[root] == -1)return;
if (nslazy[root] == 1)
{
nstree[lrt].l = nstree[lrt].r = nstree[lrt].m = (len - (len >> 1));
nstree[rrt].r = nstree[rrt].l = nstree[rrt].m = (len >> 1);
nslazy[lrt] = nslazy[root];
nslazy[rrt] = nslazy[root];
nslazy[root] = -1;
}
else
{
nstree[lrt].l = nstree[lrt].r = nstree[lrt].m = 0;
nstree[rrt].l = nstree[rrt].r = nstree[rrt].m = 0;
nslazy[lrt] = nslazy[root];
nslazy[rrt] = nslazy[root];
nslazy[root] = -1;
}
}
void build(int l, int r, int root)
{
dslazy[root] = nslazy[root] = -1;
if (l == r)
{
dstree[root].l = nstree[root].l = 1;
dstree[root].r = nstree[root].r = 1;
dstree[root].m = nstree[root].m = 1;
return;
}
int mid = (l + r) >> 1;
build(lson);
build(rson);
dspushup(root, r - l + 1);
nspushup(root, r - l + 1);
}
void dsupdate(int l, int r, int root, int lf, int rt, int val)
{
if (lf <= l && r <= rt)
{
if (val)
{
dstree[root].l = dstree[root].r = dstree[root].m = (r - l + 1);
dslazy[root] = val;
}
else
{
dstree[root].l = dstree[root].r = dstree[root].m = 0;
dslazy[root] = val;
}
return;
}
dspushdown(root, r - l + 1);
int mid = (l + r) >> 1;
if (lf <= mid)dsupdate(lson, lf, rt, val);
if (rt > mid)dsupdate(rson, lf, rt, val);
dspushup(root, r - l + 1);
}
void nsupdate(int l, int r, int root, int lf, int rt, int val)
{
if (lf <= l && r <= rt)
{
if (val)
{
nstree[root].l = nstree[root].r = nstree[root].m = (r - l + 1);
nslazy[root] = val;
}
else
{
nstree[root].l = nstree[root].r = nstree[root].m = 0;
nslazy[root] = val;
}
return;
}
nspushdown(root, r - l + 1);
int mid = (l + r) >> 1;
if (lf <= mid)nsupdate(lson, lf, rt, val);
if (rt > mid)nsupdate(rson, lf, rt, val);
nspushup(root, r - l + 1);
}
int dsquerry(int l, int r, int root, int len)
{
if (l == r)return l;
dspushdown(root, r - l + 1);
int mid = (l + r) >> 1;
if (dstree[lrt].m >= len)return dsquerry(lson, len);
else if (dstree[lrt].r + dstree[rrt].l >= len)return mid - dstree[lrt].r + 1;
else if (dstree[rrt].m >= len)return dsquerry(rson, len);
}
int nsquerry(int l, int r, int root, int len)
{
if (l == r)return l;
nspushdown(root, r - l + 1);
int mid = (l + r) >> 1;
if (nstree[lrt].m >= len)return nsquerry(lson, len);
else if (nstree[lrt].r + nstree[rrt].l >= len)return mid - nstree[lrt].r + 1;
else if (nstree[rrt].m >= len)return nsquerry(rson, len);
}
int main()
{
CASE = read();
upd(p, 1, CASE)
{
t = read(), n = read();
build(1, t, 1);
int x,y;
printf("Case %d:\n", p);
while (n--)
{
char s[20];
scanf("%s", s);
if (s[0] == 'D')
{
x = read();
if (dstree[1].m >= x)
{
int l = dsquerry(1, t, 1, x);
dsupdate(1, t, 1, l, l + x - 1, 0);
printf("%d,let's fly\n", l);
}
else printf("fly with yourself\n");
}
else if(s[0]=='N')
{
x = read();
if (dstree[1].m >= x)
{
int l = dsquerry(1, t, 1, x);
dsupdate(1, t, 1, l, l + x - 1, 0);
nsupdate(1, t, 1, l, l + x - 1, 0);
printf("%d,don't put my gezi\n", l);
}
else if (nstree[1].m >= x)
{
int l = nsquerry(1, t, 1, x);
dsupdate(1, t, 1, l, l + x - 1, 0);
nsupdate(1, t, 1, l, l + x - 1, 0);
printf("%d,don't put my gezi\n", l);
}
else printf("wait for me\n");
}
else
{
printf("I am the hope of chinese chengxuyuan!!\n");
x = read(), y = read();
dsupdate(1, t, 1, x, y, 1);
nsupdate(1, t, 1, x, y, 1);
}
}
}
return 0;
}
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