Binary Indexed Trees practice

BinaryIndexedTrees/P2352-Stars-TLE.cpp

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+// P2325-Stars.cpp : 定义控制台应用程序的入口点。
+//
+
+#include "stdafx.h"
+#include <stdio.h>
+
+#define N 15005
+int x[N], y[N];
+int res[N];
+int n;
+
+int _tmain(int argc, _TCHAR* argv[])
+{
+	int i = 0 ,j = 0, count = 0;
+	scanf("%d", &n);
+	for(i = 0; i < n; ++i) {
+		scanf("%d%d", &x[i], &y[i]);
+	}
+
+	//memset(res, 0, sizeof(res));
+
+	res[0] = 1;
+	for(i = 1; i < n; ++i) {
+		count = 0;
+		for(j = 0; j < i; ++j) {
+			if(x[j] <= x[i])
+				count++;
+		}		
+		res[count]++;
+	}
+
+	for(i = 0; i < n; ++i){
+		printf("%d\n", res[i]);
+	}
+	return 0;
+}
+

BinaryIndexedTrees/P2352-Stars.cpp

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+// P2352-Stars.cpp : 定义控制台应用程序的入口点。
+//As Stars are listed in ascending order of Y coordinate. Stars with equal Y coordinates are listed in ascending order of X coordinate. 
+//So, the stars those contribute to the star(x, y) are scanfed before (x, y). Using x coordinate is OK.  
+#include "stdafx.h"
+#include <stdio.h>
+
+#define N 15005
+#define MAXVAL 32005
+int tree[MAXVAL];
+int res[N];
+int n;
+
+void update(int x, int val) {
+	while(x <= MAXVAL) {
+		tree[x] += val;
+		x += (x & -x);
+	}
+}
+
+int read(int x) {
+	int sum = 0;
+	while(x > 0) {
+		sum += tree[x];
+		x -= (x & -x);
+	}
+	return sum;
+}
+
+int _tmain(int argc, _TCHAR* argv[])
+{
+	int i = 0 ,j = 0, count = 0;
+	int x, y;
+	scanf("%d", &n);
+	for(i = 0; i < n; ++i) {
+		scanf("%d%d", &x, &y);
+		res[read(++x)]++;
+		update(x, 1);
+	}
+
+	for(i = 0; i < n; ++i){
+		printf("%d\n", res[i]);
+	}
+	getchar();getchar();
+	return 0;
+}
+

BinaryIndexedTrees/P2481-Cows.cpp

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+// P2481-Cows.cpp : 定义控制台应用程序的入口点。
+//this problem is similar to p2352. in p2352, the secquence is order by y asc, x asc.
+//so, here we order the secquence by e desc, s asc.
+//submitted in c++
+#include "stdafx.h"
+#include <iostream>
+#include <algorithm>
+using namespace std;
+
+#define N 100005
+
+int n, tree[N], res[N];
+
+struct cow{
+	int s;
+	int e;
+	int idx;
+}cows[N];
+
+bool comp(cow a, cow b) {
+	if(a.e == b.e)
+		return a.s < b.s;
+	return a.e > b.e ;
+}
+
+int read(int x) {
+	int sum = 0;
+	while(x > 0) {
+		sum += tree[x];
+		x -= (x & -x);
+	}
+	return sum;
+}
+
+void update(int x) {
+	while(x <= n) {
+		tree[x]++;
+		x += (x & -x);
+	}
+}
+
+int _tmain(int argc, _TCHAR* argv[])
+{
+	int i, j;
+	while(scanf("%d", &n), n != 0) {
+		memset(cows, 0, sizeof(cows));
+		memset(tree, 0, sizeof(tree));
+		memset(res, 0, sizeof(res));
+
+		for(i = 0; i < n; ++i) {
+			scanf("%d%d", &cows[i].s, &cows[i].e);
+			cows[i].s++;
+			cows[i].idx = i;
+		}
+		sort(cows, cows + n, comp);  //the idea is the same as P2352
+		struct cow pre = {-1, -1};
+		int minus = 0;
+		for(i = 0; i < n; ++i) {
+			if(pre.e == cows[i].e && pre.s == cows[i].s)
+				minus++;
+			else {
+				minus = 0;
+				pre.e = cows[i].e;
+				pre.s = cows[i].s;
+			}
+			res[cows[i].idx] = read(cows[i].s)  - minus;
+			update(cows[i].s);
+		}
+		for(i = 0; i < n; i++)
+			printf("%d ", res[i]);
+		printf("\n");
+	}
+	getchar(); getchar();
+	return 0;
+}
+