[线段树]HDU5091 Beam Cannon

题意:给n, w, h  (1 < = N < = 10000, 1 < = W < = 40000, 1 < = H < = 40000)

　　w*h是可以射到的范围

　然后给n个点的坐标x, y (-20000 < = x,y < = 20000)

问 射一次 能射到几个点.

很裸的扫描线线段树

这么水的题不知为何在现场没有写啊??????

  1 #include <cstdio>
  2 #include <cstdlib>
  3 #include <cstring>
  4 #include <climits>
  5 #include <cctype>
  6 #include <cmath>
  7 #include <string>
  8 #include <sstream>
  9 #include <iostream>
 10 #include <algorithm>
 11 #include <iomanip>
 12 using namespace std;
 13 #include <queue>
 14 #include <stack>
 15 #include <vector>
 16 #include <deque>
 17 #include <set>
 18 #include <map>
 19 typedef long long LL;
 20 typedef long double LD;
 21 #define pi acos(-1.0)
 22 #define lson l, m, rt<<1
 23 #define rson m+1, r, rt<<1|1
 24 typedef pair<int, int> PI;
 25 typedef pair<int, PI> PP;
 26 #ifdef _WIN32
 27 #define LLD "%I64d"
 28 #else
 29 #define LLD "%lld"
 30 #endif
 31 //#pragma comment(linker, "/STACK:1024000000,1024000000")
 32 //LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;}
 33 //inline int read(){char ch=‘ ‘;int ans=0;while(ch<‘0‘ || ch>‘9‘)ch=getchar();while(ch<=‘9‘ && ch>=‘0‘){ans=ans*10+ch-‘0‘;ch=getchar();}return ans;}
 34 //inline void print(LL x){printf(LLD, x);puts("");}
 35 //inline void read(double &x){char c = getchar();while(c < ‘0‘) c = getchar();x = c - ‘0‘; c = getchar();while(c >= ‘0‘){x = x * 10 + (c - ‘0‘); c = getchar();}}
 36
 37 const int maxn = 100005;
 38 int X[maxn<<3];
 39 struct Seg
 40 {
 41     int h , l , r, s;
 42     Seg() {}
 43     Seg(int a,int b,int c,int d) : l(a) , r(b) , h(c) , s(d) {}
 44     bool operator < (const Seg &cmp) const
 45     {
 46         return h < cmp.h;
 47     }
 48 } ss[maxn<<3];
 49 int len[maxn<<3],cnt[maxn<<3];
 50 void PushUp(int rt)
 51 {
 52     len[rt]=max(len[rt<<1], len[rt<<1|1]);
 53 }
 54 void PushDown(int rt)
 55 {
 56     if(cnt[rt])
 57     {
 58         cnt[rt<<1]+=cnt[rt];
 59         cnt[rt<<1|1]+=cnt[rt];
 60         len[rt<<1]+=cnt[rt];
 61         len[rt<<1|1]+=cnt[rt];
 62         cnt[rt]=0;
 63     }
 64 }
 65 void Update(int L, int R, int c, int l , int r, int rt)
 66 {
 67     if(L<=l && r<=R)
 68     {
 69         cnt[rt]+=c;
 70         len[rt]+=c;
 71         return ;
 72     }
 73 //    PushDown(rt);
 74     int m=(l+r)>>1;
 75     if(L<=m)
 76         Update(L, R, c, lson);
 77     if(m<R)
 78         Update(L, R, c, rson);
 79 //    PushUp(rt);
 80     len[rt]=max(len[rt<<1], len[rt<<1|1])+cnt[rt];
 81 }
 82 int main()
 83 {
 84 #ifndef ONLINE_JUDGE
 85     freopen("in.txt", "r", stdin);
 86     freopen("out.txt", "w", stdout);
 87 #endif
 88     int n;
 89     while(~scanf("%d", &n))
 90     {
 91         if(n<0)
 92             break;
 93         int w, h;
 94         scanf("%d%d", &w, &h);
 95         int m=0;
 96         while(n--)
 97         {
 98             int a, b;
 99             scanf("%d%d", &a, &b);
100             X[m]=a;
101             ss[m++]=Seg(a, a+w, b, 1);
102             X[m]=a+w;
103             ss[m++]=Seg(a, a+w, b+h, -1);
104         }
105         sort(X, X+m);
106         sort(ss, ss+m);
107         int k=unique(X, X+m)-X;
108         int ans=0;
109         memset(len, 0, sizeof(len));
110         memset(cnt, 0, sizeof(cnt));
111         for(int i=0;i<m;i++)
112         {
113             int l=lower_bound(X, X+k, ss[i].l)-X;
114             int r=lower_bound(X, X+k, ss[i].r)-X;
115             Update(l, r, ss[i].s, 0, k-1, 1);
116             ans=max(ans, len[1]);
117         }
118         printf("%d\n", ans);
119     }
120     return 0;
121 }

HDU5091