Color the ball
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 28415 Accepted Submission(s): 13851Problem 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
C/C++(线段树):
#include <map>
#include <queue>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <climits>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;
const int MAX = 5e5 + ; int n, a, b; struct nod
{
int L, R, val;
}node[MAX]; void build(int d, int l, int r)
{
node[d].L = l, node[d].R = r, node[d].val = ;
if (l == r) return;
int mid = (l + r) >> ;
build(d << , l, mid);
build((d << ) + , mid + , r);
} void update(int d, int l, int r)
{
if (node[d].L == l && node[d].R == r)
{
node[d].val ++;
return;
}
if (node[d].L == node[d].R) return;
int mid = (node[d].L + node[d].R) >> ;
if (r <= mid)
update(d << , l, r);
else if (mid < l)
update((d << ) + , l, r);
else
{
update(d << , l, mid);
update((d << ) + , mid + , r);
}
} int query(int d, int l, int r)
{
if (node[d].L == l && node[d].R == r) return node[d].val;
if (node[d].L == node[d].R) return ;
int mid = (node[d].L + node[d].R) >> ;
if (r <= mid)
return node[d].val + query(d << , l, r);
else if (mid < l)
return node[d].val + query((d << ) + , l, r);
else
return node[d].val + query(d << , l, mid) + query((d << ) + , mid + , r);
} int main()
{
while (scanf("%d", &n), n)
{
build(, , n);
for (int i = ; i < n; ++ i)
{
scanf("%d%d", &a, &b);
update(, a, b);
}
for (int i = ; i < n; ++ i)
printf("%d ", query(, i, i));
printf("%d\n", query(, n, n));
}
return ;
}
C/C++(技巧):
#include <map>
#include <queue>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <climits>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;
const int MAX = 1e5 + ; int n, a, b; int main()
{
while (scanf("%d", &n), n)
{
int num[MAX] = {}, m = ;
for (int i = ; i <= n; ++ i)
{
scanf("%d%d", &a, &b);
num[a] ++, num[b + ] --;
}
for (int i = ; i < n; ++ i)
{
m += num[i];
printf("%d ", m);
}
printf("%d\n", m + num[n]);
}
return ;
}