Bzoj 1818 Internal White point (discretized + tree Array)

This is the weakening version of 1227.

Draw a picture or just a little proof to know that there will be no more than one change.

So we just need to count how many white dots will darken, in other words, how many white spots have black dots up or down.

The discretization of the horizontal axis, because there is no black spot in the column is meaningless, the answer also did not contribute.

Then process each row, for each row, maintain a bit that is what points will contribute, the bit will only have n times to modify, n query.

So the complexity of Time is O (Nlogn).

# include <cstdio># include<cstring># include<cstdlib># include<iostream># include<vector># include<queue># include<stack># include<map># include<Set># include<cmath># include<algorithm>using namespacestd;# define Lowbit (x) ( x)& (-x)) # define PI3.1415926535# define EPS 1e-9# define MOD9999973# define INF1000000000# define MEM (A, B) memset (A, B,sizeof(a) # define for (I,a,n) for(intI=a; i<=n; ++i) # define FO (I,a,n) for(intI=a; i<n; ++i) # define bug puts ("H"); # define LCH P<<1, l,mid# define RCH p<<1|1, mid+1, r# define MP make_pair# define PB push_backtypedef pair<int,int>pii;typedef Vector<int>vi;# pragma comment (linker,"/stack:1024000000,1024000000") typedefLong LongLL;intScan () {intres=0, flag=0; Charch; if((Ch=getchar ()) = ='-') flag=1; Else if(ch>='0'&&ch<='9') res=ch-'0';  while((Ch=getchar ()) >='0'&&ch<='9') res=res*Ten+ (ch-'0'); returnflag?-res:res;}voidOut (inta) {if(a<0) {Putchar ('-'); a=-A;} if(a>=Ten) Out (A/Ten); Putchar (A%Ten+'0');}Const intn=100005;//Code begin ...structnode{intx, y;} Node[n];vi V, vv;intCol[n], vis[n], tree[n], N;BOOLComp (Node A, Node B) {if(A.Y==B.Y)returna.x<b.x; returnA.y>b.y;}voidAddintXintVal) { while(x<=n) Tree[x]+=val, x+=lowbit (x);}intQueryintx) {    intres=0;  while(x) res+=tree[x], x-=lowbit (x); returnRes;}intMain () {LL ans=0; scanf ("%d",&N); For (I,1, N) scanf ("%d%d",&node[i].x,&node[i].y), V.PB (node[i].x);    Sort (V.begin (), V.end ()); intSiz=unique (V.begin (), V.end ())-V.begin (); Sort (Node+1, node+n+1, comp); For (I,1, N) {node[i].x=lower_bound (V.begin (), V.begin () +siz,node[i].x)-v.begin () +1; ++col[node[i].x]; }    intnow=1;  while(now<=N) {vv.clear (); VV.PB (node[now].x); ++Now ;  while(now<=n&&node[now].y==node[now-1].Y) VV.PB (node[now].x), + +Now ; Siz=vv.size (); FO (i,1, siz) ans+= (query (vv[i]-1)-query (vv[i-1])); FO (i,0, Siz) {            ++Vis[vv[i]]; if(vis[vv[i]]==1&&col[vv[i]]>1) Add (Vv[i],1); Else if(vis[vv[i]]==col[vv[i]]&&col[vv[i]]>1) Add (vv[i],-1); }} printf ("%lld\n", ans+N); return 0;}

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Bzoj 1818 Internal White point (discretized + tree Array)