poj 1990 MooFest (树状数组)
2016-10-16 18:34
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MooFest
Time Limit: 1000MS Memory Limit: 30000KDescription
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
嗯……题目先贴出来,改天再写思路……
代码如下:
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <map> #include <cmath> #include <string> #include <queue> #include <stack> #define INF 0x3f3f3f3f #define mem(x,val) memset(x,val,sizeof(x)) #define SI(x) scanf("%d",&x) #define MAX 20000+10 using namespace std; typedef long long ll; struct cow{ int v,x; }c[MAX]; int t1[MAX],t2[MAX],s[MAX]; int lowbit(int x) { return x&(-x); } void update(int pos,int v,int t[],int n) { while(pos<=n) { t[pos]+=v; pos+=lowbit(pos); } } int sum(int pos,int t[]) { int ans=0; while(pos>0) { ans+=t[pos]; pos-=lowbit(pos); } return ans; } bool cmp(cow x,cow y) { return x.v<y.v; } int main() { int n,i; while(SI(n)!=EOF) { ll msum=0; ll lessx,morex,lessnum,morenum; int max_x=0; for(i=1;i<=n;i++) scanf("%d%d",&c[i].v,&c[i].x); sort(c+1,c+n+1,cmp); mem(t1,0); mem(t2,0); mem(s,0); for(i=1;i<=n;i++) { s[i]=s[i-1]+c[i].x; max_x=max(c[i].x,max_x); } for(i=1;i<=n;i++) { update(c[i].x,c[i].x,t1,max_x); update(c[i].x,1,t2,max_x); lessnum=sum(c[i].x-1,t2); lessx=sum(c[i].x-1,t1); morenum=i-1-lessnum; morex=s[i-1]-lessx; msum+=(c[i].x*lessnum-lessx+morex-c[i].x*morenum)*c[i].v; } printf("%lld\n",msum); } return 0; }
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