I-MooFest(POJ 1990)

MooFest

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 5697		Accepted: 2481

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

Source

USACO 2004 U S Open

 

读题真是个大问题啊！！！一开始读了好久也没读懂，还是看的别人题解才明白了题意。

好吧，题意好歹明白了，可是问题又来了，怎么办？

归根到底就是一个动态改值并求和的问题，树状数组（线段树当然也可）。

我发现我的树状数组真不会用，就是用的不熟练，以前做的都直接套，我都没弄懂啥意思，一次锻炼吧。

往往成功就差那么一步啊，确是咫尺天涯。

#include <cstdio>
#include <iostream>
#include <sstream>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
#define ll long long
#define _cle(m, a) memset(m, a, sizeof(m))
#define repu(i, a, b) for(int i = a; i < b; i++)
#define MAXN  20005

struct P{
   ll v, p;
   bool operator < (const P& t) const {
     return t.v > v;
   }
}cow[MAXN];
ll c_ount[MAXN] = {0};
ll total[MAXN] = {0};
ll sum_tot[MAXN] = {0};
int n;

ll lowbit(ll x)
{
    return x & (-x);
}

void add(int x, int d, ll c[])
{
    while(x < MAXN) {
        c[x] += d;
        x += lowbit(x);
    }
}

ll Sum(ll x, ll c[])
{
    ll ret = 0;
    while(x > 0)
    {
        ret += c[x];
        x -= lowbit(x);
    }
    return ret;
}

int main()
{
    scanf("%d", &n);
    repu(i, 1, n + 1) scanf("%lld%lld", &cow[i].v, &cow[i].p);
    sort(cow + 1, cow + n + 1);
    repu(i, 1, n + 1) sum_tot[i] = sum_tot[i - 1] + cow[i].p;
    ll sum = 0, num_cow = 0, sum_total = 0;
    add(cow[1].p, 1, c_ount);
    add(cow[1].p, cow[1].p, total);
    repu(i, 2, n + 1) {
      num_cow = Sum(cow[i].p, c_ount);
      sum_total = Sum(cow[i].p, total);
      sum += cow[i].v * (num_cow * cow[i].p - sum_total
          + (sum_tot[i - 1] - sum_total - (i - 1 - num_cow) * cow[i].p));
      add(cow[i].p, 1, c_ount);
      add(cow[i].p, cow[i].p, total);
    }
    printf("%lld
", sum);
    return 0;
}