此题就是1227 的弱化版。
画个图或者稍微证明一下就能够知道，一定不会超过一次变换。
那么我们只需要统计有多少个白点会变黑，换句话说就是有多少个白点上下左右都有黑点。
离散化横坐标，因为没有黑点在的列是没有任何意义的，对答案也没有贡献。
然后处理每一行，对于每一行，维护一个BIT也就是哪些点会产生贡献，这个BIT最多只会有n次修改，n次查询。
所以时间复杂度为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 namespace std; # define lowbit(x) ((x)&(-x)) # define pi 3.1415926535 # define eps 1e-9 # define MOD 9999973 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=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_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int res=0, flag=0; char ch; if((ch=getchar())==‘-‘) flag=1; else if(ch>=‘0‘&&ch<=‘9‘) res=ch-‘0‘; while((ch=getchar())>=‘0‘&&ch<=‘9‘) res=res*10+(ch-‘0‘); return flag?-res:res; } void Out(int a) { if(a<0) {putchar(‘-‘); a=-a;} if(a>=10) Out(a/10); putchar(a%10+‘0‘); } const int N=100005; //Code begin... struct Node{int x, y;}node[N]; VI v, vv; int col[N], vis[N], tree[N], n; bool comp(Node a, Node b){ if (a.y==b.y) return a.x<b.x; return a.y>b.y; } void add(int x, int val){while (x<=n) tree[x]+=val, x+=lowbit(x);} int query(int x){ int res=0; while (x) res+=tree[x], x-=lowbit(x); return res; } int main () { 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()); int siz=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]; } int now=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; }
原文：http://www.cnblogs.com/lishiyao/p/6618699.html
