标签:bsp ble set cin stdin ref void str string
题目链接:http://codeforces.com/problemset/problem/369/E
考虑将问题转化为有多少条线段没有覆盖这些点,如果一个询问的点集是${[x1,x2,...,xn]}$那么相当于询问有多少条线段在${\left [ 1,x1 \right )\bigcup (x1,x2)\bigcup ...(x_{n-1},x_n)\bigcup(x_n,1e6+1)}$这个范围内,所以离线询问,从左至右扫过去,每到一个点就把以这个点为右端点的线段的左端点的位置加入到树状数组中,然后统计答案
1 #include<iostream> 2 #include<algorithm> 3 #include<cstdio> 4 #include<cstdlib> 5 #include<cmath> 6 #include<cstring> 7 #include<queue> 8 #include<vector> 9 #include<map> 10 using namespace std; 11 #define llg int 12 #define maxn 1000100 13 #define yyj(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout); 14 llg ans[maxn],n,m,c[maxn]; 15 vector<llg>q[maxn],belong[maxn],line[maxn]; 16 17 inline llg getint() 18 { 19 llg w=0,q=0; char c=getchar(); 20 while((c<‘0‘ || c>‘9‘) && c!=‘-‘) c=getchar(); 21 if (c==‘-‘) q=1, c=getchar(); while (c>=‘0‘ && c<=‘9‘) w=w*10+c-‘0‘, c=getchar(); 22 return q ? -w : w; 23 } 24 25 void inc(llg x) {for (;x<maxn;x+=x&-x) c[x]++;} 26 27 llg sum(llg x) {if (x==0) return 0; llg val=0; for (;x>0;x-=x&-x) val+=c[x]; return val;} 28 29 void init() 30 { 31 cin>>n>>m; 32 for (llg i=1;i<=n;i++) 33 { 34 llg l=getint(),r=getint(); 35 line[r].push_back(l); 36 } 37 for (llg i=1;i<=m;i++) 38 { 39 ans[i]=n; 40 llg k=getint(),x=0; 41 for (llg j=1;j<=k;j++) 42 { 43 llg wz=getint(); 44 q[wz].push_back(x); 45 belong[wz].push_back(i); 46 x=wz; 47 } 48 q[1000001].push_back(x); 49 belong[1000001].push_back(i); 50 } 51 } 52 53 int main() 54 { 55 yyj("ds"); 56 init(); 57 for (llg i=1;i<=1000002;i++) 58 { 59 llg w=q[i].size(); 60 for (llg j=0;j<w;j++) 61 { 62 ans[belong[i][j]]-=sum(i-1)-sum(q[i][j]); 63 } 64 w=line[i].size(); 65 for (llg j=0;j<w;j++) 66 { 67 llg x=line[i][j]; 68 inc(x); 69 } 70 } 71 for (llg i=1;i<=m;i++) printf("%d\n",ans[i]); 72 return 0; 73 }
Codeforces 396 E. Valera and Queries
标签:bsp ble set cin stdin ref void str string
原文地址:http://www.cnblogs.com/Dragon-Light/p/6805581.html