kuangbin带我飞QAQ 线段树
1. HDU1166
裸线段树点修改
#include <iostream>
#include <string.h>
#include <cstdio>
#include <queue>
#include <map>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <math.h> #define SIGMA_SIZE 26
#pragma warning ( disable : 4996 ) using namespace std;
typedef long long LL; inline LL LMax(LL a,LL b) { return a>b?a:b; }
inline LL LMin(LL a,LL b) { return a>b?b:a; }
inline int Max(int a,int b) { return a>b?a:b; }
inline int Min(int a,int b) { return a>b?b:a; }
inline int gcd( int a, int b ) { return b==?a:gcd(b,a%b); }
inline int lcm( int a, int b ) { return a/gcd(a,b)*b; } //a*b = gcd*lcm
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int mod = ;
const int maxk = 5e4+;
const int maxn = 5e4+; int N;
int num[maxn];
int sum[maxn<<];
char str[]; void init()
{
memset( num, , sizeof(num) );
memset( sum, , sizeof(sum) );
} void pushup( int rt )
{ sum[rt] = sum[rt<<]+sum[rt<<|]; } void build( int l, int r, int rt )
{
if ( l == r )
{ sum[rt] = num[l]; return; }
int m = (l+r)>>; build( l, m, rt<< );
build( m+, r, rt<<| );
pushup(rt);
} void update(int L, int C, int l, int r, int rt)
{
if( l == r )
{ sum[rt]+=C; return; } int m=(l+r)>>; if(L <= m)
update( L, C, l, m, rt<< );
else
update( L, C, m+, r, rt<<| );
pushup(rt);
} int query( int L, int R, int l, int r, int rt )
{
if ( L <= l && R >= r )
return sum[rt]; int m = (l+r)>>; int Ans = ;
if ( L <= m ) Ans += query( L, R, l, m, rt<< );
if ( R > m ) Ans += query( L, R, m+, r, rt<<| );
return Ans;
} int main()
{
int T; cin >> T;
int cnt = ;
while (T--)
{
init();
scanf("%d", &N);
for ( int i = ; i <= N; i++ )
scanf( "%d", &num[i] ); build( , N, );
printf( "Case %d:\n", cnt++ ); int x, w;
while ()
{
scanf( "%s", str );
if ( str[] == 'E' )
break; scanf( "%d %d", &x, &w );
if ( str[] == 'A' )
update( x, w, , N, );
else if ( str[] == 'S' )
update( x, -w, , N, );
else if ( str[] == 'Q' )
printf( "%d\n", query(x,w,,N,) );
}
}
return ;
}
2.HDU 1754
不知道为什么scanf读不了char，变成了“烫烫烫烫”....还是裸线段树求区间最大
#include <iostream>
#include <string.h>
#include <cstdio>
#include <queue>
#include <map>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <math.h> #define SIGMA_SIZE 26
#pragma warning ( disable : 4996 ) using namespace std;
typedef long long LL; inline LL LMax(LL a,LL b) { return a>b?a:b; }
inline LL LMin(LL a,LL b) { return a>b?b:a; }
inline int Max(int a,int b) { return a>b?a:b; }
inline int Min(int a,int b) { return a>b?b:a; }
inline int gcd( int a, int b ) { return b==?a:gcd(b,a%b); }
inline int lcm( int a, int b ) { return a/gcd(a,b)*b; } //a*b = gcd*lcm
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int mod = ;
const int maxk = 5e4+;
const int maxn = 2e5+; int N;
int num[maxn];
int mmax[maxn<<]; void pushup( int rt )
{ mmax[rt] = Max( mmax[rt<<], mmax[rt<<|] ); } void build( int l, int r, int rt )
{
if ( l == r )
{
mmax[rt] = num[l];
return;
} int m = (l+r)>>;
build( l, m, rt<< );
build( m+, r, rt<<| ); pushup(rt);
} void update( int L, int C, int l, int r, int rt )
{
if ( r == l )
{
mmax[rt] = C;
return;
} int mid = (l+r)>>;
if ( L <= mid ) update( L, C, l, mid, rt<< );
else update( L, C, mid+, r, rt<<| ); pushup(rt);
} int query( int lhs, int rhs, int l, int r, int rt )
{ if ( lhs <= l && rhs >= r )
return mmax[rt]; int mid = (l+r)>>;
int m = -; if ( lhs <= mid ) m = Max( m, query(lhs, rhs, l, mid, rt<<) );
if ( rhs > mid ) m = Max( m, query(lhs, rhs, mid+, r, rt<<|) ); return m;
} int main()
{
int N, M;
char c[];
char ch; while ( ~scanf("%d %d", &N, &M) )
{
memset( mmax, , sizeof(mmax) ); for ( int i = ; i <= N; i++ )
scanf("%d", &num[i]); build( , N, ); int x, y;
for ( int i = ; i <= M; i++ )
{
scanf("%s%d%d", &c, &x, &y); if ( c[] == 'U' )
update( x, y, , N, );
else
printf( "%d\n", query(x, y, , N, ) );
}
}
return ;
}
3.POJ 3468
简单的区间修改，但是容易错，数字全部都用long long 才行
#include <iostream>
#include <string.h>
#include <cstdio>
#include <queue>
#include <map>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <math.h> #define SIGMA_SIZE 26
#pragma warning ( disable : 4996 ) using namespace std;
typedef long long LL; inline LL LMax(LL a,LL b) { return a>b?a:b; }
inline LL LMin(LL a,LL b) { return a>b?b:a; }
inline int Max(int a,int b) { return a>b?a:b; }
inline int Min(int a,int b) { return a>b?b:a; }
inline int gcd( int a, int b ) { return b==?a:gcd(b,a%b); }
inline int lcm( int a, int b ) { return a/gcd(a,b)*b; } //a*b = gcd*lcm
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int mod = ;
const int maxk = 5e4+;
const int maxn = 1e5+; int N;
LL num[maxn], Add[maxn<<];
LL sum[maxn<<]; void pushup(LL rt) { sum[rt]=sum[rt<<]+sum[rt<<|]; } //ln是左子树数字节点的数目, rn是右子树数字节点的数目
void pushdown( LL rt, LL ln, LL rn )
{
if ( Add[rt] != )
{
//下推标记
Add[rt<<] += Add[rt];
Add[rt<<|] += Add[rt];
//修改子节点的sum
sum[rt<<] += Add[rt]*ln;
sum[rt<<|] += Add[rt]*rn; Add[rt] = ;
}
} void build( LL l, LL r, LL rt )
{
if ( l == r )
{
sum[rt] = num[l];
return;
} LL mid = (l+r)>>;
build(l, mid, rt<<);
build(mid+, r, rt<<|); pushup(rt);
} void update( LL lhs, LL rhs, LL C,LL l, LL r, LL rt )
{
if ( lhs <= l && rhs >= r )
{
sum[rt] += (LL)C*(r-l+);
Add[rt] += C;
return;
} LL mid = (l+r)>>;
pushdown( rt, mid-l+, r-mid ); if ( lhs <= mid ) update( lhs, rhs, C, l, mid, rt<< );
if ( rhs > mid ) update( lhs, rhs, C, mid+, r, rt<<| );
pushup(rt);
} LL query( int lhs, LL rhs, LL l, LL r, LL rt )
{
if ( lhs <= l && rhs >= r )
return sum[rt]; LL mid = (l+r)>>;
pushdown( rt, mid-l+, r-mid ); LL ans = ;
if ( lhs <= mid ) ans += query( lhs, rhs, l, mid, rt<< );
if ( rhs > mid ) ans += query( lhs, rhs, mid+, r, rt<<| ); return ans;
} int main()
{
int N, Q;
char str[];
cin >> N >> Q; for ( int i = ; i <= N; i++ )
scanf("%lld", &num[i]); build( , N, ); LL lhs, rhs, x;
while (Q--)
{
scanf( "%s", str );
if ( str[] == 'C' )
{ scanf("%lld %lld %lld", &lhs, &rhs, &x); update(lhs, rhs, x, , N, ); }
else
{ scanf("%lld %lld", &lhs, &rhs); printf("%lld\n", query(lhs, rhs, , N, )); }
} return ;
}
4.POJ 2528
区间染色+离散化，好题建议多做一遍，有一个坑点在于离散化后范围，你以为一共1e4对数据，所以最多2e4个点，实际上我们离散的时候如果两个区间之间相隔大于1，离散的时候是会加一个数字的，所以极限情况每对区间之间都加了一个点，所以得有3e4
个点，所以线段树数组至少得开4*3e4 = 12e4才行
#include <iostream>
#include <string.h>
#include <cstdio>
#include <queue>
#include <map>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <math.h> #define SIGMA_SIZE 26
#pragma warning ( disable : 4996 ) using namespace std;
typedef long long LL; inline LL LMax(LL a,LL b) { return a>b?a:b; }
inline LL LMin(LL a,LL b) { return a>b?b:a; }
inline int Max(int a,int b) { return a>b?a:b; }
inline int Min(int a,int b) { return a>b?b:a; }
inline int gcd( int a, int b ) { return b==?a:gcd(b,a%b); }
inline int lcm( int a, int b ) { return a/gcd(a,b)*b; } //a*b = gcd*lcm
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int mod = ;
const int maxk = 1e4+;
const int maxn = 1e6+; int N, ans;
int num[maxk<<], lisan[maxk<<];
int tree[maxk<<];
int li[maxk], ri[maxk];
bool has[maxk];
//因为是染色问题所以没有上推
void pushdown( int rt )
{
tree[rt<<] = tree[rt<<|] = tree[rt];
tree[rt] = -;
} void update( int L, int R, int C, int l, int r, int rt )
{
//因为范围已经覆盖了子树，所以直接修改标记也没关系
if ( L <= l && R >= r )
{
tree[rt] = C;
return;
} //如果有标记，必须先下推标记才能修改tree
if ( tree[rt] != - ) pushdown(rt);
int mid = (l+r)>>;
if ( L <= mid ) update( L, R, C, l, mid, rt<< );
if ( R > mid ) update( L, R, C, mid+, r, rt<<| );
//不用上推
} int binary( int x, int l, int r )
{
while ( l < r )
{
int mid = (l+r)>>; if ( lisan[mid] == x )
return mid;
if ( lisan[mid] > x )
r = mid - ;
else
l = mid + ;
}
return l;
} void query( int l, int r, int rt )
{
if ( tree[rt] != - )
{
if ( !has[tree[rt]] )
{
ans++;
has[tree[rt]] = true;
}
return;
} //如果叶子节点有海报则在上一步肯定已经被计算过了，所以直接返回就行
if ( l == r ) return; int mid = (l+r)>>;
query( l, mid, rt<< );
query( mid+, r, rt<<| );
} int main()
{
int N;
int T; cin >> T;
while (T--)
{
scanf( "%d", &N );
memset( tree, -, sizeof(tree) ); int cnt = ;
for ( int i = ; i <= N; i++ )
{
scanf( "%d %d", &li[i], &ri[i] );
num[cnt++] = li[i];
num[cnt++] = ri[i];
} sort( num+, num+cnt );
//int m = unique( num+1, num+cnt ) - num; //可以用这个去重 int tmpcnt = cnt;
for ( int i = ; i < cnt; i++ )
{
if (num[i] - num[i - ] > )
num[tmpcnt++] = num[i - ]++;
} sort( num+, num+tmpcnt ); cnt = ;
lisan[] = num[];
for ( int i = ; i < tmpcnt; i++ )
if ( num[i] != num[i-] )
lisan[cnt++] = num[i];
cnt--; for ( int i = ; i <= N; i++ )
{
int lhs = binary( li[i], , cnt );
int rhs = binary( ri[i], , cnt ); update( lhs, rhs, i, , cnt, );
} ans = ;
memset( has, , sizeof(has) );
query( , cnt, ); printf( "%d\n", ans );
}
return ;
}
5. HDU 1698
区间修改，和上一题差不多，不过我觉得顺序出反了...明显这题简单嘛
#include <iostream>
#include <string.h>
#include <cstdio>
#include <queue>
#include <map>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <math.h> #define SIGMA_SIZE 26
#pragma warning ( disable : 4996 ) using namespace std;
typedef long long LL; inline LL LMax(LL a,LL b) { return a>b?a:b; }
inline LL LMin(LL a,LL b) { return a>b?b:a; }
inline int Max(int a,int b) { return a>b?a:b; }
inline int Min(int a,int b) { return a>b?b:a; }
inline int gcd( int a, int b ) { return b==?a:gcd(b,a%b); }
inline int lcm( int a, int b ) { return a/gcd(a,b)*b; } //a*b = gcd*lcm
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int mod = ;
const int maxk = 1e5+;
const int maxn = 1e5+; int ans;
int tree[maxn<<]; //void pushup( int rt ) { sum[rt] = sum[rt<<1] + sum[rt<<1|1]; } void pushdown( int rt )
{
tree[rt<<] = tree[rt<<|] = tree[rt];
tree[rt] = -;
} void update( int L, int R, int C, int l, int r, int rt )
{
if ( L <= l && R >= r )
{
tree[rt] = C;
return;
} if ( tree[rt] != - ) pushdown(rt); int mid = (l+r)>>;
if ( L <= mid ) update( L, R, C, l, mid, rt<< );
if ( R > mid ) update( L, R, C, mid+, r, rt<<| );
//不用上推
} void query( int l, int r, int rt )
{
//特判叶子节点
if ( l == r )
{
if (tree[rt] == -) ans++;
else
ans += tree[rt];
return;
} if ( tree[rt]!=- )
{
ans += tree[rt]*(r-l+);
return;
} int mid = (l+r)>>;
query(l, mid, rt<<);
query(mid+, r, rt<<|);
} int main()
{
int N, Q;
int T, cnt = ; cin >> T; while (T--)
{
ans = ;
memset( tree, -, sizeof(tree) ); scanf( "%d", &N ); scanf( "%d", &Q ); int lhs, rhs, z;
for ( int i = ; i <= Q; i++ )
{
scanf( "%d %d %d", &lhs, &rhs, &z );
update( lhs, rhs, z, , N, );
} query( , N, );
printf( "Case %d: The total value of the hook is %d.\n", cnt++, ans );
}
return ;
}
6. POJ 3264
水题，求区间最大减最小
#include <iostream>
#include <string.h>
#include <cstdio>
#include <queue>
#include <map>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <math.h> #define SIGMA_SIZE 26
#define lson rt<<1
#define rson rt<<1|1
#pragma warning ( disable : 4996 ) using namespace std;
typedef long long LL;
inline LL LMax(LL a,LL b) { return a>b?a:b; }
inline LL LMin(LL a,LL b) { return a>b?b:a; }
inline int Max(int a,int b) { return a>b?a:b; }
inline int Min(int a,int b) { return a>b?b:a; }
inline int gcd( int a, int b ) { return b==?a:gcd(b,a%b); }
inline int lcm( int a, int b ) { return a/gcd(a,b)*b; } //a*b = gcd*lcm
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod = ;
const int inf = 0x3f3f3f3f;
const int maxk = 1e5+;
const int maxn = 2e5+; int hei[maxn];
int mmin[maxn<<], mmax[maxn<<]; void pushup( int rt )
{
mmin[rt] = Min( mmin[rt<<], mmin[rt<<|] );
mmax[rt] = Max( mmax[rt<<], mmax[rt<<|] );
} void build( int l, int r, int rt )
{
if ( l == r )
{
mmin[rt] = hei[l];
mmax[rt] = hei[l];
return;
} int mid = (l+r)>>;
build( l, mid, rt<< );
build( mid+, r, rt<<| );
pushup(rt);
} int findmin( int L, int R, int l, int r, int rt )
{
if ( L <= l && R >= r )
return mmin[rt]; int mid = (l+r)>>;
int mi = inf; if ( L <= mid ) mi = Min( mi, findmin(L,R,l,mid,rt<<) );
if ( R > mid ) mi = Min( mi, findmin(L,R,mid+,r,rt<<|) );
return mi;
} int findmax( int L, int R, int l, int r, int rt )
{
if ( L <= l && R >= r )
return mmax[rt]; int mid = (l+r)>>;
int ma = -; if ( L <= mid ) ma = Max( ma, findmax(L,R,l,mid,rt<<) );
if ( R > mid ) ma = Max( ma, findmax(L,R,mid+,r,rt<<|) );
return ma;
} int main()
{
int N, Q;
cin >> N >> Q; for ( int i = ; i <= N; i++ )
scanf( "%d", &hei[i] ); build( , N, ); int l, r;
while (Q--)
{
scanf( "%d%d", &l, &r );
printf("%d\n", findmax(l,r,,N,) - findmin(l,r,,N,) );
} return ;
}
7.HDU 1540
区间合并
求某个点左右两边的最长连续长度
恶心的题目，明明是多组数据写的却只有一组,各种WA的我又去看了kuangbin的代码结果又被误导了,,,看的我一愣一愣的最后发现还是自己写的好使...太坑了
#include <iostream>
#include <string.h>
#include <cstdio>
#include <queue>
#include <map>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <math.h> #define SIGMA_SIZE 26
#define lson rt<<1
#define rson rt<<1|1
#pragma warning ( disable : 4996 ) using namespace std;
typedef long long LL;
inline LL LMax(LL a,LL b) { return a>b?a:b; }
inline LL LMin(LL a,LL b) { return a>b?b:a; }
inline int Max(int a,int b) { return a>b?a:b; }
inline int Min(int a,int b) { return a>b?b:a; }
inline int gcd( int a, int b ) { return b==?a:gcd(b,a%b); }
inline int lcm( int a, int b ) { return a/gcd(a,b)*b; } //a*b = gcd*lcm
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod = ;
const int inf = 0x3f3f3f3f;
const int maxk = 1e5+;
const int maxn = 5e4+; //rl代表从右到左最大连续村庄数，ll表示从左到右最大，ml表示该区间最大
struct qujian {
int l; //区间长度
int ll, rl, ml;
}qj[maxn<<];
int sta[maxn], top;
bool dead[maxn]; void pushup( int rt )
{
int m = qj[lson].rl+qj[rson].ll; if (qj[lson].ml==qj[lson].l) qj[rt].ll = qj[lson].ll+qj[rson].ll;
else qj[rt].ll = qj[lson].ll; if (qj[rson].ml==qj[rson].l) qj[rt].rl = qj[lson].rl+qj[rson].rl;
else qj[rt].rl = qj[rson].rl; qj[rt].ml = Max( qj[rson].ml, qj[lson].ml );
qj[rt].ml = Max( m, qj[rt].ml );
} void build( int l, int r, int rt )
{
if ( l == r )
{
qj[rt].l = ;
qj[rt].ll = ;
qj[rt].ml = ;
qj[rt].rl = ;
return;
} int mid = (l+r)>>;
build( l, mid, rt<< );
build( mid+, r, rt<<| );
qj[rt].l = qj[lson].l + qj[rson].l;
pushup(rt);
} //删除是false， 恢复是true
void update( int L, int l, int r, int rt, bool flag )
{
if ( l == r )
{
qj[rt].ll = qj[rt].rl = qj[rt].ml = flag?:;
return;
} int mid = (l+r)>>;
if ( L <= mid ) update( L, l, mid, lson, flag );
else update( L, mid+, r, rson, flag );
pushup(rt);
} int query( int L, int l, int r, int rt )
{
if ( l==r || qj[rt].ml==qj[rt].l || qj[rt].ml== )
return qj[rt].ml; int mid = (l+r)>>;
//如果在左子树，则lson的rl如果把查询点包括了，则要加上rson
if ( L <= mid )
{
if ( qj[lson].rl- >= mid-L )
return qj[lson].rl + qj[rson].ll;
return query( L, l, mid, lson );
}
else //同上
{
if ( qj[rson].ll- >= L-mid- )
return qj[rson].ll + qj[lson].rl;
return query( L, mid+, r, rson );
}
} int main()
{
int n, m;
while ( ~scanf("%d%d", &n, &m) )
{
top = ;
memset( dead, , sizeof(dead) );
build(,n,); char str[];
int x;
while (m--)
{
scanf( "%s", str );
if ( str[] == 'D' )
{
scanf("%d",&x);
sta[top++] = x;
if (!dead[x])
update( x, , n, , false );
}
else if (str[] == 'R')
{
if ( top )
update(sta[--top], , n, , true);
}
else
{
scanf("%d",&x);
if (dead[x]) printf( "0\n" );
else
printf( "%d\n", query(x,,n,) );
}
}
}
return ;
}
8.HDU 3974
dfs建树，这是一道需要比较深入理解线段树操作的题目，WA了很多次是因为dfs的时候写成了start[i] = cnt++，这样如果到了叶子节点，那么start[i] 和end[i]就不等了,,,坑爹的是这种错误样例根本测不出来...
#include <iostream>
#include <string.h>
#include <cstdio>
#include <queue>
#include <map>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <math.h> #define SIGMA_SIZE 26
#define lson rt<<1
#define rson rt<<1|1
#pragma warning ( disable : 4996 ) using namespace std;
typedef long long LL;
inline LL LMax(LL a,LL b) { return a>b?a:b; }
inline LL LMin(LL a,LL b) { return a>b?b:a; }
inline int Max(int a,int b) { return a>b?a:b; }
inline int Min(int a,int b) { return a>b?b:a; }
inline int gcd( int a, int b ) { return b==?a:gcd(b,a%b); }
inline int lcm( int a, int b ) { return a/gcd(a,b)*b; } //a*b = gcd*lcm
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod = ;
const int inf = 0x3f3f3f3f;
const int maxk = 1e5+;
const int maxn = 5e4+; struct edge {
int to, next;
}e[maxn]; int cnt;
int st[maxn], ed[maxn];
int linjie[maxn], tree[maxn<<], change[maxn<<];
bool vis[maxn]; void addedge( int u, int v )
{ e[cnt].to = v; e[cnt].next = linjie[u]; linjie[u] = cnt++; } void init()
{
cnt = ;
memset( linjie, -, sizeof(linjie) );
memset( vis, , sizeof(vis) );
memset( tree, -, sizeof(tree) );
memset( change, -, sizeof(change) );
} //妙，妙啊！
void dfs( int u )
{
st[u] = ++cnt;
for ( int i = linjie[u]; i+; i = e[i].next )
dfs(e[i].to);
ed[u] = cnt;
} void pushdown( int rt )
{
if ( change[rt] != - )
{
int tmp = change[rt];
change[lson] = change[rson] = tmp;
tree[lson] = tree[rson] = tmp; change[rt] = -;
}
} void update( int L, int R, int C, int l, int r, int rt )
{
if ( L <= l && R >= r )
{
tree[rt] = C;
change[rt] = C;
return;
} int mid = (l+r)>>;
pushdown(rt); if ( L <= mid ) update( L, R, C, l, mid, lson );
if ( R > mid ) update( L, R, C, mid+, r, rson );
} int query( int L, int l, int r, int rt )
{
if ( L == l && L == r ) // L <= l && L >= r
return tree[rt]; pushdown(rt);
int tmp = -, mid = (l+r)>>; if ( L <= mid ) tmp = query( L, l, mid, lson );
else tmp = query( L, mid+, r, rson ); return tmp;
} int main()
{
freopen("F:\\cpp\\vs\\test\\Debug\\test.txt","r",stdin); int T; cin >> T; int N, Q, tot = ;
char str[]; while (T--)
{
init();
cin >> N; int x, y;
for ( int i = ; i < N; i++ )
{
scanf( "%d %d", &x, &y );
vis[x] = true;
addedge( y, x ); //从y到x
} cnt = ;
for ( int i = ; i <= N; i++ )
if ( !vis[i] )
{ dfs(i); break; } cin >> Q;
printf("Case #%d:\n", tot++);
for ( int i = ; i <= Q; i++ )
{
scanf( "%s", str );
if ( str[] == 'C' )
{
scanf( "%d", &x );
printf( "%d\n", query(st[x], , cnt, ) );
}
else
{
scanf( "%d %d", &x, &y );
update( st[x], ed[x], y, , cnt, );
}
} }
return ;
}
9 HDU 4614
给N个瓶子，第一个操作，从第A个瓶子开始往里插x朵插花，如果一个瓶子里面没有花则放一朵花进去，如果有则找下一个直到最后一个瓶子，如果花没用完则舍弃掉，输出放第一朵花和最后一朵花的瓶子号
第二个操作，舍弃l~r区间内瓶子里的花，输出舍弃的花的数目
我是用数组1代表瓶子内无花，0代表有花，则区间和就代表没有花的瓶子个数，然后第一个操作就二分出起始瓶子和结尾瓶子，第二个操作直接区间长度-区间和
PS：十个二分九个错
#include <iostream>
#include <string.h>
#include <cstdio>
#include <queue>
#include <map>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <math.h> #define SIGMA_SIZE 26
#define lson rt<<1
#define rson rt<<1|1
#pragma warning ( disable : 4996 ) using namespace std;
typedef long long LL;
inline LL LMax(LL a,LL b) { return a>b?a:b; }
inline LL LMin(LL a,LL b) { return a>b?b:a; }
inline int Max(int a,int b) { return a>b?a:b; }
inline int Min(int a,int b) { return a>b?b:a; }
inline int gcd( int a, int b ) { return b==?a:gcd(b,a%b); }
inline int lcm( int a, int b ) { return a/gcd(a,b)*b; } //a*b = gcd*lcm
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL mod = ;
const int inf = 0x3f3f3f3f;
const int maxk = 1e5+;
const int maxn = 5e4+; int sum[maxn<<];
int chag[maxn<<]; void init()
{
memset( sum, , sizeof(sum) );
memset( chag, -, sizeof(chag) );
} void pushup( int rt ) { sum[rt] = sum[lson] + sum[rson]; } void build( int l, int r, int rt )
{
if ( l == r )
{
sum[rt] = ;
return;
} int mid = (l+r)>>;
build( l, mid, lson );
build( mid+, r, rson );
pushup(rt);
} //注意这个下推，只有两种操作，一个全部置0，一个全部置1
//置1代表这个瓶子是空的，置0表示有花
void pushdown( int rt, int ln, int rn )
{
if ( chag[rt] != - )
{
chag[lson] = chag[rson] = chag[rt];
sum[lson] = ln*chag[lson];
sum[rson] = rn*chag[rson];
chag[rt] = -;
}
} void update( int L, int R, int C, int l, int r, int rt )
{
if ( L <= l && R >= r )
{
sum[rt] = (r-l+)*C;
chag[rt] = C;
return;
} int mid = (l+r)>>;
pushdown( rt, mid-l+, r-mid ); if ( L <= mid ) update( L, R, C, l, mid, lson );
if ( R > mid ) update( L, R, C, mid+, r, rson );
pushup(rt);
} int query( int L, int R, int l, int r, int rt )
{
if ( L <= l && R >= r )
return sum[rt]; int mid = (l+r)>>;
pushdown( rt, mid-l+, r-mid ); int ans = ;
if ( L <= mid ) ans += query( L, R, l, mid, lson );
if ( R > mid ) ans += query( L, R, mid+, r, rson ); return ans;
} int bin( int st, int ed, int val, bool ok )
{
int mid, tmp;
int lhs = st, rhs = ed;
while ( lhs <= rhs )
{
mid = (lhs+rhs)>>;
tmp = query( st, mid, , ed, );
if ( (ok?(val <= tmp):(val < tmp)) )
rhs = mid-;
else
lhs = mid+;
}
return lhs;
} int main()
{
//freopen("F:\\cpp\\test.txt","r",stdin); int T; cin >> T; int N, Q, k, x, y;
while (T--)
{
init();
cin >> N >> Q;
build(, N, ); int lpos, rpos;
while (Q--)
{
scanf("%d%d%d", &k, &x, &y);
if ( k == )
{
//t代表空瓶数量
int t = query(x + , N, , N, );
if ( t == )
{
printf("Can not put any one.\n");
continue;
}
else if ( t < y ) //如果该段空花盆数目不够放
rpos = bin( x+, N, t, true ); //找出第一个空瓶数量等于t的位置
else
rpos = bin( x+, N, y, true ); //如果够放直接二分终止位置 //二分起始位置,查找第一个不等于0的数
lpos = bin( x+, N, , false ); printf( "%d %d\n", lpos-, rpos- ); //注意对齐位置
update( lpos, rpos, , , N, );
}
else
{
int len = y-x+, tmp = query(x+, y+, , N, );
printf("%d\n", len-tmp);
update( x+, y+, , , N, );
}
}
printf("\n");
}
return ;
}
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