# [树状数组] 树状数组基础练习题题解（三）

POJ-1990 MooFest

Time Limit: 1000MS
Memory Limit: 30000K

## 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 US Open

#include
#include
#include
#include
#include
#include
#define LL long long

using namespace std;

LL x = 0;
LL p = 1;
char ch = getchar();

while(ch < '0' || ch > '9'){
if(ch == '-')
p = -1;
ch = getchar();
}

while('0' <= ch && ch <= '9'){
x = x*10 + ch - '0';
ch = getchar();
}

return x*p;
}

const int MAXN = 20000;

struct Data{
LL v;
LL x;
};

Data input[MAXN+10];

inline bool cmp(Data p,Data q){
return p.v
 
 
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