hdu 题目1556 Color the ball（线段树）

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Color the ball

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5853 Accepted Submission(s): 3117

Problem Description

N个气球排成一排，从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的“小飞鸽"牌电动车从气球a开始到气球b依次给每个气球涂一次颜色。但是N次以后lele已经忘记了第I个气球已经涂过几次颜色了，你能帮他算出每个气球被涂过几次颜色吗？

Input

每个测试实例第一行为一个整数N,(N <= 100000).接下来的N行，每行包括2个整数a b(1 <= a <= b <= N)。
当N = 0，输入结束。

Output

每个测试实例输出一行，包括N个整数，第I个数代表第I个气球总共被涂色的次数。

Sample Input

31 12 23 331 11 21 30

Sample Output

1 1 13 2 1

#define N 400000#include <iostream>#include <stdio.h>using namespace std;typedef struct interval_tree{    int lside,rside,cnt;}interval_tree;interval_tree tree[N];int Pnt[N],k;void build_tree(int l,int r,int i){            tree[i].lside = l;        tree[i].rside = r;        tree[i].cnt = 0;        if(l==r) return;        int mid = (tree[i].rside+tree[i].lside)/2;        build_tree(l,mid,2*i);        build_tree(mid+1,r,2*i+1);    }void count(int i){if(tree[i].lside == tree[i].rside)     {Pnt[k++] = tree[i].cnt;return;}tree[i*2].cnt += tree[i].cnt;tree[i*2+1].cnt += tree[i].cnt;count(2*i);count(2*i+1);}void change(int a,int b,int i){ if(a<=tree[i].lside && b>=tree[i].rside)      {            tree[i].cnt++;        return;    }    int mid =(tree[i].lside+tree[i].rside)/2;    if(b <= mid )     change(a,b,2*i);    else if(a > mid )    change(a,b,2*i+1);    else     {        change(a,mid,2*i);        change(mid+1,b,2*i+1);    }}int main(){    int i,n,a,b;    while(scanf("%d",&n),n)    {        build_tree(1,n,1);        while(n--)        {            scanf("%d%d",&a,&b);            change(a,b,1);        }        k=0;        count(1);        for(i=0;i<k-1;i++)        printf("%d ",Pnt[i]);        printf("%d\n",Pnt[k-1]);    }    return 0;}