1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 #define Max 32010 5 int star[Max]; 6 int level[Max]; 7 int Lowbit(int x); 8 int Sum(int k); 9 void Update(int k); 10 int main() 11 { 12 //freopen("D:\\acm.txt","r",stdin); 13 int xNum = 0, yNum = 0; 14 int lineNum = 0; 15 scanf("%d",&lineNum); 16 for (int i = 0; i < lineNum; ++i){ 17 scanf("%d%d", &xNum, &yNum); 18 level[Sum(xNum + 1)]++;//xNum要加1或者一個正整數,避免爲0,否則會報錯 19 Update(xNum + 1); 20 } 21 for (int i = 0; i < lineNum; ++i) 22 printf("%d\n",level[i]); 23 return 0; 24 } 25 int Lowbit(int x){ 26 return x & (-x);//返回最後一個1的位置開始的數 27 } 28 int Sum(int k){ //前k項和 29 int judge = 0; 30 while (k){ 31 judge = judge + star[k]; 32 k = k - Lowbit(k);// 33 } 34 return judge; 35 } 36 void Update(int k){//修改第k項,更新之後的內容 37 while (k < Max){ 38 star[k]++; 39 k = k + Lowbit(k); 40 } 41 }
附上線段樹解決方案(不是我寫的),留作以後學習。現在基本是看懂代碼,理解思想,真正自己是想不到的,繼續努力!
http://blog.chinaunix.net/uid-22263887-id-1778936.html