bzoj4378 [POI2015]Logistyka
对于询问(a,b)，设X=(数列中小于等于b的数字和),Y=b*(数列中大于b的数字的个数),若X+Y>=a*b，则询问可行，否则不可行，实现的话用两个树状数组维护一下即可
代码
1 #include<cstdio> 2 #include<algorithm> 3 #include<set> 4 #define fi first 5 #define sc second 6 #define mp make_pair 7 #define N 2000100 8 using namespace std; 9 long long c1[N],c2[N],e[N],a[N],b[N]; 10 int typ[N],n,i,m,tot,u,v[N]; 11 char ch; 12 int ef(int x) 13 { 14 int l=1,r=n,m; 15 while (l<=r) 16 { 17 m=(l+r)>>1; 18 if (e[m]>x) r=m-1;else l=m+1; 19 } 20 return r; 21 } 22 int lowbit(int x) 23 { 24 return x&-x; 25 } 26 void cc(int x,int w,long long ww) 27 { 28 while (x<=n) 29 { 30 c1[x]+=w; 31 c2[x]+=ww; 32 x+=lowbit(x); 33 } 34 } 35 pair<int,long long> sum(int x) 36 { 37 int ans1=0; 38 long long ans2=0; 39 while (x) 40 { 41 ans1+=c1[x]; 42 ans2+=c2[x]; 43 x-=lowbit(x); 44 } 45 return mp(ans1,ans2); 46 } 47 int main() 48 { 49 scanf("%d%d",&n,&m); 50 tot++;e[tot]=0; 51 for (i=1;i<=m;i++) 52 { 53 scanf(" %c %lld%lld",&ch,&a[i],&b[i]); 54 if (ch=='U') 55 { 56 tot++;e[tot]=b[i]; 57 typ[i]=0; 58 } 59 else typ[i]=1; 60 } 61 sort(e+1,e+1+tot);e[0]=-1;n=0; 62 for (i=1;i<=tot;i++) 63 if (e[i]!=e[i-1]) e[++n]=e[i]; 64 cc(1,n,n*e[1]); 65 for (i=1;i<=m;i++) 66 { 67 if (!typ[i]) 68 { 69 u=ef(v[a[i]]); 70 cc(u,-1,-v[a[i]]); 71 v[a[i]]=b[i]; 72 u=ef(v[a[i]]); 73 cc(u,1,v[a[i]]); 74 } 75 else 76 { 77 u=ef(b[i]); 78 pair<int,long long> ans2=sum(u); 79 long long v1=n-ans2.fi; 80 long long v2=ans2.sc; 81 if (v1*b[i]+v2>=a[i]*b[i]) 82 printf("TAK\n"); 83 else 84 printf("NIE\n"); 85 } 86 } 87 }
