hdu4417 (Offline operation + tree-like array)

Test instructions

Given an array of length n, a query of M times, how many of the numbers are smaller than h for each query [a, b] interval?

Since the values of N and M range from 0 to 10 for 5, the direct answer will time out, so consider reading all the query operations first, then answering the number in the [a, b] interval that is smaller than H.

Summation is similar to the method of finding positive ordinal numbers.

Writing can be inserted into the search, you can also look at the side of the plug, while the side of the simple;

But this problem from the beginning WA to the tail, because the Add function less dozen equals number, slap himself.

Side by side search:

#include <iostream>#include<cstdio>#include<time.h>#include<stdlib.h>#include<cstring>#include<algorithm>#defineMAXN 110000#defineLL intusing namespacestd;structnode{LL Value,id;};structquerynode{LL l,r,h=-1, CNT;}; Node A[maxn];querynode QUE[MAXN]; LL C[MAXN]; LL n,m; LL ANS[MAXN];voidinit () {memset (A,0,sizeof(a)); Memset (c,0,sizeof(c)); memset (ans,0,sizeof(ANS));}BOOLCMP1 (node A,node b) {returna.value<B.value; }BOOLCMP2 (Querynode a,querynode b) {returna.h<B.H;}intInline Lowbit (intx) {    returnX & (-x);}voidInline Add (intx) {     while(x <=N) {C[x]++; X+=lowbit (x); }}intInline sum (intx) {    ints =0;  while(X >0) {s+=C[x]; X-=lowbit (x); }    returns;}voidsolve () {intt=1;  for(intI=1; i<=n;i++)    {            if(Que[t]. h==-1)            {                 Break; }             while(Que[t]. H<a[i].value && Que[t]. h!=-1) {ans[que[t].cnt]=sum (Que[t]. R)-sum (Que[t]. L-1); T++;            } add (A[i].id);  while(a[i].value==a[i+1].value && i<=N) {i++;            Add (a[i].id); }             while(Que[t]. H==a[i].value && Que[t]. h!=-1) {ans[que[t].cnt]=sum (Que[t]. R)-sum (Que[t]. L-1); T++; }    }     for(inti=t;i<=m;i++) {ans[que[i].cnt]=sum (Que[i]. R)-sum (Que[i]. L-1); }}intMain () {//freopen ("Test.txt", "R", stdin);    intT; scanf ("%d",&t); intCase=1;  for(intCase=1; case<=t; case++) {scanf ("%d%d",&n,&M);      Init ();  for(intI=1; i<=n;i++) {scanf ("%d",&a[i].value); A[i].id=i; }       for(intI=1; i<=m;i++) {scanf ("%d%d%d", &que[i]. L,&que[i]. r,&Que[i].         H); Que[i]. L++; Que[i]. R++; Que[i].cnt=i; } sort (A+1, a+n+1, CMP1); Sort (que+1, que+m+1, CMP2);       Solve (); printf ("Case %d:\n", case);  for(inti =1; I <= M; ++i) {printf ("%d\n", Ans[i]); }    }    return 0;}

Side-by-side interpolation:

#include <iostream>#include<cstdio>#include<time.h>#include<stdlib.h>#include<cstring>#include<algorithm>#defineMAXN 110000#defineLL intusing namespacestd;structnode{LL Value,id;};structquerynode{LL l,r,h,cnt;}; Node A[maxn];querynode QUE[MAXN]; LL C[MAXN]; LL n,m; LL ANS[MAXN];voidinit () {memset (A,0,sizeof(a)); Memset (c,0,sizeof(c)); memset (ans,0,sizeof(ans));  for(intI=0; i<maxn;i++) {Que[i]. H=-1; }}BOOLCMP1 (node A,node b) {returna.value<B.value; }BOOLCMP2 (Querynode a,querynode b) {returna.h<B.H;}intInline Lowbit (intx) {    returnX & (-x);}voidInline Add (intx) {     while(x <=N) {C[x]++; X+=lowbit (x); }}intInline sum (intx) {    ints =0;  while(X >0) {s+=C[x]; X-=lowbit (x); }    returns;}voidsolve1 () { for(intAski =0, DITJ =0; Aski < M; ++Aski) {            while(Ditj < N && Que[aski]. H >=a[ditj].value)              {Add (a[ditj].id); DITJ++; } ans[que[aski].cnt]= SUM (Que[aski]. R)-sum (Que[aski]. L1); }}intMain () {//freopen ("Test.txt", "R", stdin);    intT; scanf ("%d",&t); intCase=1;  for(intCase=1; case<=t; case++) {scanf ("%d%d",&n,&l);      Init ();  for(intI=0; i<n;i++) {scanf ("%d",&a[i].value); A[i].id=i+1; }       for(intI=0; i<m;i++) {scanf ("%d%d%d", &que[i]. L,&que[i]. r,&Que[i].         H); Que[i]. L++; Que[i]. R++; Que[i].cnt=i+1; } sort (A,a+N,CMP1); Sort (Que,que+M,CMP2);       Solve1 (); printf ("Case %d:\n", case);  for(inti =1; I <= M; ++i) {printf ("%d\n", Ans[i]); }    }    return 0;}

hdu4417 (Offline operation + tree-like array)