4237: 稻草人
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4237: 稻草人
https://www.lydsy.com/JudgeOnline/problem.php?id=4237
分析:
CDQ分治+单调栈。
首先按照x排序,每次分治,考虑左边对右边的贡献。CDQ的过程中,按照y从大到小排序。
代码:
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #include<cmath> 5 #include<iostream> 6 #include<cctype> 7 #include<set> 8 #include<vector> 9 #include<queue> 10 #include<map> 11 #include<stack> 12 #define fi(s) freopen(s,"r",stdin); 13 #define fo(s) freopen(s,"w",stdout); 14 using namespace std; 15 typedef long long LL; 16 17 inline int read() { 18 int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch==‘-‘)f=-1; 19 for(;isdigit(ch);ch=getchar())x=x*10+ch-‘0‘;return x*f; 20 } 21 22 const int N = 200005; 23 24 int n; 25 struct Node{ 26 int x, y; 27 Node() {} 28 Node(int _x,int _y) { x = _x, y = _y; } 29 }A[N], B[N], sk1[N], sk2[N]; 30 int top1, top2; 31 LL Ans; 32 33 bool operator < (const Node &A, const Node &B) { 34 return A.x < B.x; 35 } 36 37 int find(int k) { 38 int L = 1, R = top2 + 1, res = top2 + 1; 39 while (L <= R) { 40 int mid = (L + R) >> 1; 41 if (sk2[mid].y < k) res = mid, R = mid - 1; // sk中y是递减的 42 else L = mid + 1; 43 } 44 return res; 45 } 46 47 void CDQ(int l,int r) { 48 if (l >= r) return ; 49 int mid = (l + r) >> 1; 50 CDQ(l, mid); CDQ(mid + 1, r); 51 52 top1 = 0, top2 = 0; 53 int i = l, j = mid + 1, k; 54 for (i=l; i<=mid; ++i) { 55 while (top1 && sk1[top1].x < A[i].x) top1--; // x单调递减 56 sk1[++top1] = A[i]; 57 while (j<=r && A[j].y >= A[i].y) { 58 while (top2 && sk2[top2].x > A[j].x) top2 --; // x单调递增 59 sk2[++top2] = A[j ++]; 60 } 61 if (top1 == 1) Ans += top2; 62 else Ans += top2 - find(sk1[top1-1].y) + 1; 63 } 64 i = l, j = mid + 1, k = l; 65 while (i <= mid && j <= r) { 66 if (A[i].y > A[j].y) B[k++] = A[i ++]; 67 else B[k++] = A[j ++]; 68 } 69 while (i <= mid) B[k ++] = A[i ++]; 70 while (j <= r) B[k ++] = A[j ++]; 71 for (i=l; i<=r; ++i) A[i] = B[i]; 72 } 73 74 int main() { 75 n = read(); 76 for (int i=1; i<=n; ++i) { 77 int x = read(), y = read(); 78 A[i] = Node(x, y); 79 } 80 sort(A + 1, A + n + 1); 81 CDQ(1, n); 82 cout << Ans; 83 return 0; 84 }
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