題干：

Recently, the γ galaxies broke out Star Wars. Each planet is warring for resources. In the Star Wars, Planet X is under attack by other planets. Now, a large wave of enemy spaceships is approaching. There is a very large Beam Cannon on the Planet X, and it is very powerful, which can destroy all the spaceships in its attack range in a second. However, it takes a long time to fill the energy of the Beam Cannon after each shot. So, you should make sure each shot can destroy the enemy spaceships as many as possible.?

To simplify the problem, the Beam Cannon can shot at any area in the space, and the attack area is rectangular. The rectangle parallels to the coordinate axes and cannot rotate. It can only move horizontally or vertically. The enemy spaceship in the space can be considered as a point projected to the attack plane. If the point is in the rectangular attack area of the Beam Cannon(including border), the spaceship will be destroyed.

Input

Input contains multiple test cases. Each test case contains three integers N(1<=N<=10000, the number of enemy spaceships), W(1<=W<=40000, the width of the Beam Cannon’s attack area), H(1<=H<=40000, the height of the Beam Cannon’s attack area) in the first line, and then N lines follow. Each line contains two integers x,y (-20000<=x,y<=20000, the coordinates of an enemy spaceship).?

A test case starting with a negative integer terminates the input and this test case should not to be processed.

Output

Output the maximum number of enemy spaceships the Beam Cannon can destroy in a single shot for each case.

Sample Input

2 3 4 0 1 1 0 3 1 1 -1 0 0 1 1 0 -1

Sample Output

2 2

題目大意：

? 給你一個矩形的寬度和長度，然后給你一些點的坐標，讓你輸出這個矩形最多能覆蓋多少個點。

解題報告：

? 類似掃描線的思想，將一條邊拆成兩條邊，權值分別為1和-1。

AC代碼1：（其實這里的.f，用laz比較合適）

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair using namespace std; const int MAX = 2e5 + 5; struct Node {int x,y;int f;bool operator < (const Node &a)const{if(y==a.y) return f<a.f;return y<a.y;} } node[MAX]; struct TREE {int l,r;int f,val; } tree[400005 * 16];//??????????????????? void pushdown(int cur) {if(tree[cur].f == 0) return;int tmp = tree[cur].f;tree[cur].f = 0;tree[cur*2].f += tmp;tree[cur*2+1].f += tmp;tree[cur*2].val += tmp;tree[cur*2+1].val += tmp; } void build(int l,int r,int cur) {tree[cur].l=l,tree[cur].r=r;tree[cur].f = tree[cur].val = 0;if(l == r) return;int m = (l+r)>>1;build(l,m,cur*2);build(m+1,r,cur*2+1); } void update(int pl,int pr,int val,int cur) {if(pl <= tree[cur].l && pr >= tree[cur].r) {//pushup(cur);tree[cur].f += val;tree[cur].val += val;return ;}pushdown(cur);if(tree[cur*2].r >= pl) update(pl,pr,val,cur*2);if(tree[cur*2+1].l <= pr) update(pl,pr,val,cur*2+1);//pushup(cur);tree[cur].val = max(tree[cur*2].val,tree[cur*2+1].val); } int main() {int n,W,H;while(cin>>n) {if(n == -1) break;cin>>W>>H;for(int a,b,i = 1; i<=2*n; i+=2) {scanf("%d%d",&a,&b);a+=20005;node[i].x=a;node[i+1].x=a;node[i].y=b;node[i+1].y=b+H+1;node[i].f=1;node[i+1].f=-1;}sort(node+1,node+2*n+1);build(1,10005*8,1);int ans = 0;for(int i = 1; i<=2*n; i++) {update(node[i].x,node[i].x + W,node[i].f,1);ans = max(ans,tree[1].val);}printf("%d\n",ans); }return 0 ; }

AC代碼3：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair using namespace std; const int MAX = 2e5 + 5; struct Node {int x,y;int f;bool operator < (const Node &a)const{return x<a.x;} } node[MAX]; struct TREE {int l,r;int f,val; } tree[400005 * 16];//??????????????????? void pushdown(int cur) {if(tree[cur].f == 0) return;int tmp = tree[cur].f;tree[cur].f = 0;tree[cur*2].f += tmp;tree[cur*2+1].f += tmp;tree[cur*2].val += tmp;tree[cur*2+1].val += tmp; } void build(int l,int r,int cur) {tree[cur].l=l,tree[cur].r=r;tree[cur].f = tree[cur].val = 0;if(l == r) return;int m = (l+r)>>1;build(l,m,cur*2);build(m+1,r,cur*2+1); } void update(int pl,int pr,int val,int cur) {if(pl <= tree[cur].l && pr >= tree[cur].r) {//pushup(cur);tree[cur].f += val;tree[cur].val += val;return ;}pushdown(cur);if(tree[cur*2].r >= pl) update(pl,pr,val,cur*2);if(tree[cur*2+1].l <= pr) update(pl,pr,val,cur*2+1);//pushup(cur);tree[cur].val = max(tree[cur*2].val,tree[cur*2+1].val); } int main() {int n,W,H;while(cin>>n) {if(n == -1) break;cin>>W>>H;for(int a,b,i = 1; i<=2*n; i+=2) {scanf("%d%d",&a,&b);b+=20000;node[i].x=a;node[i+1].x=a+W+1;node[i].y=b;node[i+1].y=b;node[i].f=1;node[i+1].f=-1;}sort(node+1,node+2*n+1);build(0,10005*8,1);int ans = 0;for(int i = 1; i<=2*n; i++) {update(node[i].y,node[i].y + H,node[i].f,1);ans = max(ans,tree[1].val);}printf("%d\n",ans); }return 0 ; }

AC代碼4：（其實排序來說最標準的應該是這樣）

struct Node {int x,y;int f;bool operator < (const Node &a)const{if(x==a.x) return f<a.f;return x<a.x;} } node[MAX];

?

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