HDU 1566 Color the ball（树状数组or线段树）

Color the ball

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11387    Accepted Submission(s): 5680

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

 

Author

8600

 

Source

HDU 2006-12 Programming Contest

 

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    思路：题意很容易懂只是用一般的办法会超时，所以需要用到一些省时间的算法，树状数组和线段树就恰恰符合这个条件

树状数组

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<string.h>#include<stdlib.h>using namespace std;int a[100010];int n;int lowbit(int x){     return x&(-x);}void build(int x,int y){    while(x<=n)    {        a[x] += y;        x += lowbit(x);    }}int sum(int x)//求和{    int s=0;    while(x>0)    {        s=s+a[x];        x=x-lowbit(x);    }    return s;}int main(){    while(scanf("%d",&n)!=EOF)    {        if(n == 0)        {            break;        }        int x,y;        memset(a,0,sizeof(a));        for(int i=1;i<=n;i++)        {            scanf("%d%d",&x,&y);            build(x,1);            build(y+1,-1);        }        for(int i=1;i<=n;i++)        {            if(i == n)            {                printf("%d\n",sum(i));            }            else            {                printf("%d ",sum(i));            }        }    }    return 0;}

线段树：

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<string.h>#include<stdlib.h>using namespace std;const int maxn = 100010;struct node{    int l;    int r;    int lz;    int ans;} q[maxn<<4];int n;void pushdown(int rt,int lr){    if(q[rt].lz)    {        q[rt<<1].ans += (lr-(lr>>1))*q[rt].lz;        q[rt<<1|1].ans += (lr>>1)*q[rt].lz;        q[rt<<1].lz += q[rt].lz;        q[rt<<1|1].lz += q[rt].lz;        q[rt].lz = 0;    }}void build(int l,int r,int rt){    q[rt].l = l;    q[rt].r = r;    q[rt].ans = 0;    q[rt].lz = 0;    if(l == r)    {        return ;    }    int mid = (l+r)>>1;    build(l,mid,rt<<1);    build(mid+1,r,rt<<1|1);    q[rt].ans = q[rt<<1].ans + q[rt<<1|1].ans;}void updata(int ll,int rr,int k,int l,int r,int rt){    if(ll>r || rr<l)    {        return ;    }    if(ll<=l && rr>=r)    {        q[rt].ans = q[rt].ans + (r-l+1)*k;        q[rt].lz += k;        return ;    }    pushdown(rt,r-l+1);    int mid = (l+r)>>1;    if(mid>=ll)    {        updata(ll,rr,k,l,mid,rt<<1);    }    if(rr>mid)    {        updata(ll,rr,k,mid+1,r,rt<<1|1);    }    q[rt].ans = q[rt<<1].ans + q[rt<<1|1].ans;}void qurry(int l,int r,int rt){    if(l == r)    {        if(l == n)        {            printf("%d\n",q[rt].ans);        }        else        {            printf("%d ",q[rt].ans);        }        return ;    }    pushdown(rt,r-l+1);    int mid = (l+r)>>1;    qurry(l,mid,rt<<1);    qurry(mid+1,r,rt<<1|1);}int main(){    while(scanf("%d",&n)!=EOF)    {        if(n == 0)        {            break;        }        build(1,n,1);        int x,y;        for(int i=0; i<n; i++)        {            scanf("%d%d",&x,&y);            updata(x,y,1,1,n,1);        }        qurry(1,n,1);    }    return 0;}