POJ 1990 :MooFest 线段树
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大概我们可以先把他们按照v从大到小排个序
然后来两个树状数组记录一下他前面有多少头牛,以及牛的横坐标的前缀和
然后……就能算了
然后算完一个删除一个……就行了
#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>#define N 20000+5#define M 80000+5using namespace std;long long tr[M],sum[M];struct ask{ long long x; long long val; int pos;}q[N];inline int read(){ int x=0,f=1;char ch = getchar(); while(ch < '0' || ch > '9'){if(ch == '-')f=-1;ch = getchar();} while(ch >='0' && ch <='9'){x=(x<<1)+(x<<3)+ch-'0';ch = getchar();} return x*f;}bool cmp1(const ask &a,const ask &b){ return a.x < b.x;}bool cmp2(const ask&a,const ask&b){ return a.val > b.val;}inline void updata(int k){ tr[k] = tr[k<<1]+tr[k<<1|1]; sum[k] = sum[k<<1]+sum[k<<1|1];}void build(int k,int l,int r){ if(l==r){sum[k]=1,tr[k]=q[l].x;return;} int mid = (l+r)>>1; build(k<<1,l,mid); build(k<<1|1,mid+1,r); updata(k);}inline void change(int k,int l,int r,int pos){ if(l==r) { tr[k] = sum[k] = 0; return; } int mid = (l+r)>>1; if(pos<=mid)change(k<<1,l,mid,pos); else change(k<<1|1,mid+1,r,pos); updata(k);}int asksum(int k,int l,int r,int x,int y){ if(l==x && r==y) return tr[k]; int mid = (l+r)>>1; if(y<=mid)return asksum(k<<1,l,mid,x,y); else if(x>mid)return asksum(k<<1|1,mid+1,r,x,y); else return asksum(k<<1,l,mid,x,mid)+asksum(k<<1|1,mid+1,r,mid+1,y);}int asknum(int k,int l,int r,int x,int y){ if(l==x && r==y) return sum[k]; int mid = (l+r)>>1; if(y<=mid)return asknum(k<<1,l,mid,x,y); else if(x>mid)return asknum(k<<1|1,mid+1,r,x,y); else return asknum(k<<1,l,mid,x,mid)+asknum(k<<1|1,mid+1,r,mid+1,y);}int main(){ int n = read(); for(int i=1;i<=n;++i) q[i].val=read(),q[i].x=read(); sort(q+1,q+n+1,cmp1); for(int i=1;i<=n;++i) q[i].pos = i; build(1,1,n); sort(q+1,q+n+1,cmp2); long long ans = 0; for(int i=1;i<=n;++i) { int num = asknum(1,1,n,1,q[i].pos); long long bsum = asksum(1,1,n,1,q[i].pos); num--; bsum-=q[i].x; long long tmp1 = num * q[i].x - bsum; long long tmp2 = tr[1]-bsum-q[i].x; long long tmp3 = sum[1]-num-1; tmp3 *= q[i].x; tmp2 -= tmp3; tmp1 += tmp2 ; ans += q[i].val * tmp1; change(1,1,n,q[i].pos); } cout<<ans<<endl;}
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