POJ-1990 MooFest (树状数组 入门题)

MooFest

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 8515		Accepted: 3854

Description

Every year, Farmer John's N (1 <= N <= 20,000) cows attend "MooFest",a social gathering of cows from around the world. MooFest involves a variety of events including haybale stacking, fence jumping, pin the tail on the farmer, and of course, mooing. When the
cows all stand in line for a particular event, they moo so loudly that the roar is practically deafening. After participating in this event year after year, some of the cows have in fact lost a bit of their hearing. 

Each cow i has an associated "hearing" threshold v(i) (in the range 1..20,000). If a cow moos to cow i, she must use a volume of at least v(i) times the distance between the two cows in order to be heard by cow i. If two cows i and j wish to converse, they
must speak at a volume level equal to the distance between them times max(v(i),v(j)). 

Suppose each of the N cows is standing in a straight line (each cow at some unique x coordinate in the range 1..20,000), and every pair of cows is carrying on a conversation using the smallest possible volume. 

Compute the sum of all the volumes produced by all N(N-1)/2 pairs of mooing cows. 

Input

* Line 1: A single integer, N 

* Lines 2..N+1: Two integers: the volume threshold and x coordinate for a cow. Line 2 represents the first cow; line 3 represents the second cow; and so on. No two cows will stand at the same location. 

Output

* Line 1: A single line with a single integer that is the sum of all the volumes of the conversing cows. 

Sample Input

4
3 1
2 5
2 6
4 3

Sample Output
57

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define maxn 20005
int c[maxn], c1[maxn];
struct tree{
int pos, val;
}a[maxn];
bool mysort(tree e1, tree e2){
return e1.pos < e2.pos;
}
inline int lowbit(int x){
return x & (-x);
}
void add(int c[], int x, int v){
while(x < maxn){
c[x] += v;
x += lowbit(<
4000
/span>x);
}
}
int query(int c[], int x){
int ans = 0;
while(x){
ans += c[x];
x -= lowbit(x);
}
return ans;
}
int main(){
int n;
scanf("%d", &n);
for(int i = 1; i <= n; ++i){
scanf("%d %d", &a[i].val, &a[i].pos);
}
sort(a + 1, a + 1 + n, mysort);
memset(c, 0, sizeof(c));
memset(c1, 0, sizeof(c1));
long long ans = 0, pos;
int cnt;
for(int i = 1; i <= n; ++i){
cnt = query(c, a[i].val - 1);
add(c, a[i].val, 1);
pos = query(c1, a[i].val - 1);
add(c1, a[i].val, a[i].pos);
ans += 1LL * a[i].val * (cnt * a[i].pos - pos);
}
memset(c, 0, sizeof(c));
memset(c1, 0, sizeof(c1));
for(int i = n; i >= 1; --i){
cnt = query(c, a[i].val);
add(c, a[i].val, 1);
pos = query(c1, a[i].val);
add(c1, a[i].val, a[i].pos);
ans += 1LL * a[i].val * (pos - cnt * a[i].pos);
}
printf("%lld\n", ans);
}

/*
题意：2e4头牛在一条主线上，坐标范围2e4，声量范围2e4，两头牛之间交流的代价是两头牛之间的距离乘上两头牛声量
的较大值，求所有牛相互交流的代价和。

思路：暴力就是n方了，怎么优化呢，我们可以在声量上做文章。对于一头牛，它和其余声量小于它的牛交流，一定是用它
的音量，这样我们只要统计声量小于它的牛的距离和，然后减减乘乘就行了。坐标范围不大，显然可以树状数组搞之。
对于所有牛我们按坐标排序，然后统计每头牛前面有多少只牛声量比它小，然后再对距离求和，这都是树状数组简单应用了。
正反各搞一遍。
*/