I-MooFest(POJ 1990)

时间：2022-06-20 14:49:03
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

 
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] = {};

ll total[MAXN] = {};

ll sum_tot[MAXN] = {};

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 = ;

    while(x > )

    {

        ret += c[x];

        x -= lowbit(x);

    }

    return ret;

}

int main()

{

    scanf("%d", &n);

    repu(i, , n + ) scanf("%lld%lld", &cow[i].v, &cow[i].p);

    sort(cow + , cow + n + );

    repu(i, , n + ) sum_tot[i] = sum_tot[i - ] + cow[i].p;

    ll sum = , num_cow = , sum_total = ;

    add(cow[].p, , c_ount);

    add(cow[].p, cow[].p, total);

    repu(i, , n + ) {

      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 - ] - sum_total - (i -  - num_cow) * cow[i].p));

      add(cow[i].p, , c_ount);

      add(cow[i].p, cow[i].p, total);

    }

    printf("%lld\n", sum);

    return ;

}

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