Color the ball
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 28152 Accepted Submission(s): 13695
Problem Description
N個氣球排成一排,從左到右依次編號爲1,2,3....N.每次給定2個整數a b(a <= b),lele便爲騎上他的“小飛鴿"牌電動車從氣球a開始到氣球b依次給每個氣球塗一次顏色。但是N次以後lele已經忘記了第I個氣球已經塗過幾次顏色了,你能幫他算出每個氣球被塗過幾次顏色嗎?
Input
每個測試實例第一行爲一個整數N,(N <= 100000).接下來的N行,每行包括2個整數a b(1 <= a <= b <= N)。
當N = 0,輸入結束。
Output
每個測試實例輸出一行,包括N個整數,第I個數代表第I個氣球總共被塗色的次數。
Sample Input
3
1 1
2 2
3 3
3
1 1
1 2
1 3
0
Sample Output
1 1 1
3 2 1
線段樹修改區間問題
#include <stdio.h>
#define MAXN 100000
struct segNode{
int val;
int l,r;
} tr[MAXN<<2];
void build( int id , int s , int e ){
tr[id].val = 0;
tr[id].l = s;
tr[id].r = e;
if( s==e )
return ;
int mid = ( s+e )/2;
build( id*2,s,mid );
build( id*2+1,mid+1,e );
}
void updata( int id , int s , int e ){
if( tr[id].l==s && tr[id].r==e ){
tr[id].val++;
return ;
}
if( tr[id].l == tr[id].r )
return ;
int mid = ( tr[id].l + tr[id].r ) / 2;
if( mid>=e )
updata( id*2,s,e );
else if( mid<s )
updata( id*2+1,s,e );
else{
updata( id*2,s,mid );
updata( id*2+1,mid+1,e );
}
}
int query( int id , int l , int r ){
if( tr[id].l==l && tr[id].r==r )
return tr[id].val;
if( tr[id].l == tr[id].r )
return 0;
int mid = ( tr[id].l + tr[id].r ) / 2;
if( mid>=r )
return tr[id].val + query( id*2,l,r );
if( mid<l )
return tr[id].val + query( id*2+1,l,r );
return tr[id].val + query( id*2,l,mid ) + query( id*2+1,mid+1,r );
}
int main(){
int n,s,e,i,sum;
while( ~scanf( "%d",&n ) && n ){
build( 1,1,n );
for( i=1 ; i<=n ; i++ ){
scanf( "%d%d",&s,&e );
updata( 1,s,e );
}
for( i=1 ; i<=n ; i++ ){
sum = query( 1,i,i );
printf( "%d",sum );
if( i!=n )
printf( " " );
}
printf( "\n" );
}
}