poj 1990 MooFest （两个树状数组）

MooFest

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 4582		Accepted: 1913

DescriptionEvery 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, theymust 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

参考题解http://blog.sina.com.cn/s/blog_6635898a0100q0lk.html

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <stack>#include <vector>#include <map>#include <cmath>#include <cstdlib>#define L(rt) (rt<<1)#define R(rt) (rt<<1|1)#define ll __int64#define eps 1e-6using namespace std;const int INF = 1000000000;const int maxn = 20005;struct node{int v,x;}cow[maxn];int n;bool cmp(node a, node b){return a.x < b.x;}int lowbit(int x){return x & (- x);}void add(int x, int d, ll *a){for(int i = x; i < maxn; i += lowbit(i))a[i] += d;}ll sum(int x, ll *a){int ret = 0;for(int i = x; i > 0; i -= lowbit(i))ret += a[i];return ret;}int main(){ll num[maxn], dis[maxn];while(~scanf("%d", &n)){for(int i = 0; i < n; i++)scanf("%d%d", &cow[i].v, &cow[i].x);sort(cow, cow + n, cmp);memset(num, 0, sizeof(num));memset(dis, 0, sizeof(dis));ll ans = 0;for(int i = 0; i < n; i++){ans += (sum(cow[i].v - 1, num) * cow[i].x-sum(cow[i].v - 1, dis)) * cow[i].v;add(cow[i].v, 1, num);add(cow[i].v, cow[i].x, dis);}memset(num, 0, sizeof(num));memset(dis, 0, sizeof(dis));for(int i = n - 1; i >= 0; i--){ans += (sum(cow[i].v, dis) - sum(cow[i].v, num) * cow[i].x) * cow[i].v;add(cow[i].v, 1, num);add(cow[i].v, cow[i].x, dis);}printf("%I64d\n", ans);}}

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