HDU 1556 区间更新的扫线法
Color the ball
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4460 Accepted Submission(s): 2392
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
3 1 1 2 2 3 3 3 1 1 1 2 1 3 0
Sample Output
1 1 1 3 2 1
Author
8600
解题思路: 给定区间 a,b,只需记 tree[a]+=1;tree[b+1]-=1;就行;所有区间录入后,tree[i]=tree[i]+tree[i-1];即可;
代码:
#include <iostream>
#include <stdio.h>
#include <cstring>
using namespace std;
const int maxn=100005;
int tree[maxn];
int num=0;
int main()
{
int n,a,b;
while(cin>>n&&n)
{
memset(tree,0,sizeof(tree));
for(int i=1;i<=n;i++)
{
scanf("%d%d",&a,&b);
tree[a]+=1;
tree[b+1]+=-1;
}
tree[0]=0;
int tmp=0;
for(int i=1;i<n;i++)
{
tmp=tmp+tree[i];
printf("%d ",tmp);
}
tmp=tmp+tree[n];
printf("%d\n",tmp);
}
//cout << "Hello world!" << endl;
return 0;
}
#include <stdio.h>
#include <cstring>
using namespace std;
const int maxn=100005;
int tree[maxn];
int num=0;
int main()
{
int n,a,b;
while(cin>>n&&n)
{
memset(tree,0,sizeof(tree));
for(int i=1;i<=n;i++)
{
scanf("%d%d",&a,&b);
tree[a]+=1;
tree[b+1]+=-1;
}
tree[0]=0;
int tmp=0;
for(int i=1;i<n;i++)
{
tmp=tmp+tree[i];
printf("%d ",tmp);
}
tmp=tmp+tree[n];
printf("%d\n",tmp);
}
//cout << "Hello world!" << endl;
return 0;
}