树状数组——POJ 2481
对应POJ题目：点击打开链接
Cows
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 12588 | Accepted: 4174 |
Description
Farmer John's cows have discovered that the clover growing along the ridge of the hill (which we can think of as a one-dimensional number line) in his field is particularly good.
Farmer John has N cows (we number the cows from 1 to N). Each of Farmer John's N cows has a range of clover that she particularly likes (these ranges might overlap). The ranges are defined by a closed interval [S,E].
But some cows are strong and some are weak. Given two cows: cow i and cow j, their favourite clover range is [Si, Ei] and [Sj, Ej]. If Si <= Sj and Ej <= Ei and Ei - Si > Ej - Sj, we say that cow i is stronger than cow j.
For each cow, how many cows are stronger than her? Farmer John needs your help!
Farmer John has N cows (we number the cows from 1 to N). Each of Farmer John's N cows has a range of clover that she particularly likes (these ranges might overlap). The ranges are defined by a closed interval [S,E].
But some cows are strong and some are weak. Given two cows: cow i and cow j, their favourite clover range is [Si, Ei] and [Sj, Ej]. If Si <= Sj and Ej <= Ei and Ei - Si > Ej - Sj, we say that cow i is stronger than cow j.
For each cow, how many cows are stronger than her? Farmer John needs your help!
Input
The input contains multiple test cases.
For each test case, the first line is an integer N (1 <= N <= 10 5), which is the number of cows. Then come N lines, the i-th of which contains two integers: S and E(0 <= S < E <= 10 5) specifying the start end location respectively of a range preferred by some cow. Locations are given as distance from the start of the ridge.
The end of the input contains a single 0.
For each test case, the first line is an integer N (1 <= N <= 10 5), which is the number of cows. Then come N lines, the i-th of which contains two integers: S and E(0 <= S < E <= 10 5) specifying the start end location respectively of a range preferred by some cow. Locations are given as distance from the start of the ridge.
The end of the input contains a single 0.
Output
For each test case, output one line containing n space-separated integers, the i-th of which specifying the number of cows that are stronger than cow
i.
Sample Input
3 1 2 0 3 3 4 0
Sample Output
1 0 0
Hint
Huge input and output,scanf and printf is recommended.
1.按E降序排序，E相等就按S升序排序，这样比i牛强的一定在i牛前面。
排序前：
3 6
4 9
1 8
2 7
4 6
4 9
排序后：
4 9
1 8
2 7
3 6
4 6
这样，会树状数组就能做了
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
const int MAXN=100000+10;
const int INF=1<<25;
using namespace std;
int c[MAXN];
int ans[MAXN];
struct edge
{
int s,e,k;
}E[MAXN];
bool cmp(edge c1, edge c2)
{
if(c1.e != c2.e) return c1.e>c2.e;
return c1.s<c2.s;
}
int lowbit(int x)
{
return x&(-x);
}
void updata(int x, int num)
{
while(x<=MAXN)
{
c[x]+=num;
x+=lowbit(x);
}
}
int sum(int x)
{
int res=0;
while(x>0)
{
res+=c[x];
x-=lowbit(x);
}
return res;
}
int main()
{
//freopen("in.txt","r",stdin);
int n;
while(scanf("%d", &n), n)
{
memset(ans,0,sizeof(ans));
memset(c,0,sizeof(c));
int i,j;
for(i=1; i<=n; i++){
scanf("%d%d", &E[i].s, &E[i].e);
E[i].k=i;
}
sort(E+1, E+n+1, cmp);
updata(E[1].s+1, 1);
for(i=2; i<=n; i++){
if(E[i].s==E[i-1].s && E[i].e==E[i-1].e)
ans[E[i].k]=ans[E[i-1].k];
else
ans[E[i].k]=sum(E[i].s+1);
updata(E[i].s+1, 1);
}
for(i=1; i<n; i++) printf("%d ", ans[i]);
printf("%d\n", ans[n]);
}
return 0;
}