源自：http://www.cnblogs.com/jiaohuang/archive/2010/09/23/1833440.html
1投影线段总的长度
//x轴上有若干条线段，求线段覆盖的总长度。
代码
1 #include < cstdio > 2 #include < cstring > 3 4 using namespace std; 5 6 #define maxn 2001 7 8 typedef struct { 9 int cover; 10 }TreeNode; 11 TreeNode tree[maxn]; 12 13 void Insert( int p, int l, int r, int a, int b){ 14 int m; 15 if (tree[p].cover == 0 ) 16 { 17 m = (l + r) / 2 ; 18 if (a == l && b == r) 19 { 20 tree[p].cover = 1 ; 21 } 22 else if (b <= m) 23 Insert(p * 2 , l , m , a , b); 24 else if (a >= m) 25 Insert(p * 2 + 1 , m , r , a, b); 26 else 27 { 28 Insert(p * 2 , l , m , a , m); 29 Insert(p * 2 + 1 , m , r , m , b); 30 } 31 } 32 } 33 34 int Count( int p , int l, int r){ 35 if (tree[p].cover == 1 ) 36 return r - l; 37 else if (r - l <= 1 ) return 0 ; 38 else return Count(p * 2 , l ,(l + r) / 2 ) + Count(p * 2 + 1 , (l + r) / 2 , r); 39 } 40 41 42 int main(){ 43 int nl, nr; 44 memset(tree, 0 , sizeof (tree)); 45 while ( 1 ) 46 { 47 scanf( " %d %d " , & nl, & nr); 48 if (nl == 0 && nr == 0 ) break ; 49 Insert( 1 , 1 , 1000 , nl , nr); 50 } 51 printf( " %d\n " , Count( 1 , 1 , 1000 ) ); 52 return 0 ; 53 } 54
2投影线段的颜色种类
//x轴上有若干条不同线段，将它们依次染上不同的颜色，问最后能看到多少种不同的颜色？//（后染的颜色会覆盖原先的颜色）
//我们可以这样规定：x轴初始是颜色0，第一条线段染颜色1，第二条线段染颜色2，以此//类推.
代码
1
#include
<
cstdio
>
2
#include
<
cstring
>
3
4
using
namespace
std;
5
6
#define
maxn 2001
7
8
typedef
struct
{
9
int
cover;
10
}TreeNode;
11
TreeNode tree[maxn];
12
int
flag[maxn];
13
14
void
Insert(
int
p,
int
l,
int
r,
int
a,
int
b,
int
color){
15
int
m;
16
if
(tree[p].cover
!=
color)
17
{
18
m
=
(l
+
r)
/
2
;
19
if
(a
==
l
&&
b
==
r)
20
tree[p].cover
=
color;
21
else
22
{
23
if
(tree[p].cover
>=
0
)
24
{
25
tree[p
*
2
].cover
=
tree[p].cover;
26
tree[p
*
2
+
1
].cover
=
tree[p].cover;
27
tree[p].cover
=
-
1
;
28
}
29
if
(b
<=
m)
30
Insert(p
*
2
, l , m , a , b , color);
31
else
if
(a
>=
m)
32
Insert(p
*
2
+
1
, m , r , a, b , color);
33
else
34
{
35
Insert(p
*
2
, l , m , a , m , color);
36
Insert(p
*
2
+
1
, m , r , m , b , color);
37
}
38
}
39
}
//
if
40
}
41
42
void
Count(
int
p ,
int
l,
int
r){
43
if
(tree[p].cover
>=
0
)
44
{
45
flag[tree[p].cover ]
=
1
;
46
return
;
47
}
48
else
if
(r
-
l
==
1
)
return
;
49
else
50
{
51
Count(p
*
2
, l ,(l
+
r)
/
2
) ;
52
Count(p
*
2
+
1
, (l
+
r)
/
2
, r);
53
}
54
}
55
56
57
int
main(){
58
int
nl, nr , ncolor;
59
int
ans
=
0
;
60
memset(tree,
0
,
sizeof
(tree));
61
memset(flag ,
0
,
sizeof
(flag));
62
while
(
1
)
63
{
64
scanf(
"
%d %d
"
,
&
nl,
&
nr);
65
if
(nl
==
0
&&
nr
==
0
)
break
;
66
scanf(
"
%d
"
,
&
ncolor);
67
Insert(
1
,
1
,
1000
, nl , nr , ncolor);
68
}
69
Count(
1
,
1
,
1000
);
70
for
(
int
i
=
0
; i
<
maxn ; i
++
)ans
+=
flag[i];
71
printf(
"
%d\n
"
, ans
-
1
);
//
0
72
return
0
;
73
}
74
3 x轴被不同颜色分成多少段
//线段的颜色可以相同。连续的相同颜色被视作一段
代码
1 #include < cstdio > 2 #include < cstring > 3 4 using namespace std; 5 6 #define maxn 2001 7 8 typedef struct { 9 int cover; 10 }TreeNode; 11 TreeNode tree[maxn]; 12 13 void Insert( int p, int l, int r, int a, int b, int color){ 14 int m; 15 if (tree[p].cover != color) 16 { 17 m = (l + r) / 2 ; 18 if (a == l && b == r) 19 tree[p].cover = color; 20 else 21 { 22 if (tree[p].cover >= 0 ) 23 { 24 tree[p * 2 ].cover = tree[p].cover; 25 tree[p * 2 + 1 ].cover = tree[p].cover; 26 tree[p].cover = - 1 ; 27 } 28 if (b <= m) 29 Insert(p * 2 , l , m , a , b , color); 30 else if (a >= m) 31 Insert(p * 2 + 1 , m , r , a, b , color); 32 else 33 { 34 Insert(p * 2 , l , m , a , m , color); 35 Insert(p * 2 + 1 , m , r , m , b , color); 36 } 37 } 38 } // if 39 } 40 41 int Count( int p , int l, int r , int & lc , int & rc){ 42 int result , tl ,tr ; 43 if (tree[p].cover >= 0 ) 44 { 45 lc = tree[p].cover; 46 rc = tree[p].cover; 47 if (tree[p].cover > 0 ) return 1 ; 48 else return 0 ; 49 } 50 else if (r - l > 1 ) 51 { 52 result = Count(p * 2 , l , (l + r) / 2 , lc , tl) + Count(p * 2 + 1 , (l + r) / 2 ,r ,tr ,rc); 53 if (tl == tr && tl > 0 ) // the left and right sides are of the same color 54 result -= 1 ; 55 return result; 56 } 57 } 58 59 60 int main(){ 61 int nl , nr , ncolor; 62 int lc, rc; 63 int ans = 0 ; 64 memset(tree, 0 , sizeof (tree)); 65 while ( 1 ) 66 { 67 scanf( " %d %d " , & nl, & nr); 68 if (nl == 0 && nr == 0 ) break ; 69 scanf( " %d " , & ncolor); 70 Insert( 1 , 1 , 1000 , nl , nr , ncolor); 71 } 72 printf( " %d\n " , Count( 1 , 1 , 1000 , rc , lc)); // 0 73 return 0 ; 74 } 75
4 x轴上有若干条不同线段，问某个单位区间[x,x+1]上重叠了多少条线段？
代码
1 #include < cstdio > 2 #include < cstring > 3 4 using namespace std; 5 6 #define maxn 2001 7 8 typedef struct { 9 int cover , count; 10 }TreeNode; 11 TreeNode tree[maxn]; 12 13 void Insert( int p, int l, int r, int a, int b){ 14 int m; 15 m = (l + r) / 2 ; 16 if (a == l && b == r) 17 tree[p].count += 1 ; 18 else 19 { 20 if (b <= m) 21 Insert(p * 2 , l , m , a , b ); 22 else if (a >= m) 23 Insert(p * 2 + 1 , m , r , a, b ); 24 else 25 { 26 Insert(p * 2 , l , m , a , m ); 27 Insert(p * 2 + 1 , m , r , m , b ); 28 } 29 } 30 } 31 32 int Count( int p , int l, int r , int a , int b){ 33 int result , m; 34 result = 0 ; 35 while ( 1 ) 36 { 37 m = (l + r) / 2 ; 38 result = result + tree[p].count; 39 if (a == l && b == r) break ; 40 if (b <= m)r = m , p = 2 * p; 41 else if ( a >= m)l = m , p = 2 * p + 1 ; 42 } 43 return result; 44 } 45 46 47 int main(){ 48 int nl, nr , ncolor; 49 int lc, rc; 50 int ans = 0 ; 51 memset(tree, 0 , sizeof (tree)); 52 while ( 1 ) 53 { 54 scanf( " %d %d " , & nl, & nr); 55 if (nl == 0 && nr == 0 ) break ; 56 Insert( 1 , 1 , 1000 , nl , nr ); 57 } 58 printf( " %d\n " , Count( 1 , 1 , 1000 , 9 , 10 )); // 0 59 return 0 ; 60 } 61
5树状数组
//求[a , b]区间内的和 ， 可以动态改变数组的值
代码
1 #include < cstdio > 2 #include < cstring > 3 4 5 #define MAXN 10000+1 // the max number 6 int num[MAXN]; // the first number must be num[1]!! 7 8 int Lowbit( int x) // (x --> 2^k) 9 { 10 return x & (x ^ (x - 1 )); 11 } 12 13 int Sum( int x) // x>=0 14 { 15 int result; 16 result = 0 ; 17 while (x > 0 ) 18 { 19 result = result + num[x]; 20 x = x - Lowbit( x ); 21 } 22 return result; 23 } 24 25 void Modify( int p , int delta){ 26 while ( p <= MAXN) 27 { 28 num[p] = num[p] + delta; 29 p = p + Lowbit(p); 30 } 31 } 32 33 int main(){ 34 int nt , nplace; 35 memset(num , 0 , sizeof (num)); 36 while ( 1 ) 37 { 38 scanf( " %d %d " , & nplace, & nt); 39 if (nplace == 0 && nt == 0 ) break ; 40 Modify(nplace , nt); 41 } 42 43 return 0 ; 44 } 45
6二维树状数组
代码
1
#include
<
cstdio
>
2
#include
<
cstring
>
3
4
5
#define
MAXN 1000+1
//
the max number
6
int
num[MAXN][MAXN];
//
the first number must be num[1][1]!!
7
8
int
Lowbit(
int
x)
//
(x --> 2^k)
9
{
10
return
x
&
(x
^
(x
-
1
));
11
}
12
13
int
Sum(
int
x ,
int
y)
//
x>=0
14
{
15
int
result , t;
16
result
=
0
;
17
while
(x
>
0
)
18
{
19
t
=
y;
20
while
(t
>
0
)
21
{
22
result
=
result
+
num[x][t];
23
t
=
t
-
Lowbit(t);
24
}
25
x
=
x
-
Lowbit(x);
26
}
27
return
result;
28
}
29
30
void
Modify(
int
x ,
int
y ,
int
delta){
31
int
t;
32
while
( x
<
MAXN)
33
{
34
t
=
y;
35
while
(t
<
MAXN)
36
{
37
num[x][t]
=
num[x][t]
+
delta;
38
t
=
t
+
Lowbit(t);
39
}
40
x
=
x
+
Lowbit( x );
41
}
42
}
43
44
int
main(){
45
int
nt , nx , ny;
46
memset(num ,
0
,
sizeof
(num));
47
while
(
1
)
48
{
49
scanf(
"
%d %d %d
"
,
&
nx ,
&
ny,
&
nt);
50
if
(nx
==
0
&&
ny
==
0
)
break
;
51
Modify(nx , ny , nt);
52
}
53
printf(
"
%d \n
"
, Sum(
100
,
100
));
54
return
0
;
55
}
56
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