水题四道 3-30
题目1:： http://poj.org/problem?id=2833
大意:：n个数,去掉最大的n1个和最小的n2个数,(n1+n2 < n)，求剩下的平均值。
题解：节省内存开销，使用两个堆维护，用最小堆维护最大的n1个数，用最大堆维护最小的n2个数，插满时，只需比较两个堆的front，再决定是否插入
代码：
#include <stdio.h>#include <string.h>const int INF = 0x7fffffff;template<typename _Type>class MaxHeap
{
private: _Type data[20]; int size; int cur;public: MaxHeap(int _n):size(_n),cur(0)
{ memset(data,0,sizeof(data)); data[0] = INF;
} ~MaxHeap() { } void clear() { memset(data,0,sizeof(data)); cur = 0; data[0] = INF; } void push(_Type _value) { if(isFull()) { return ; } cur ++; int i; for(i = cur; data[i/2] < _value;i/=2) { data[i] = data[i/2]; } data[i] = _value; } void pop() { if(isEmpty()) return ; int lastElement = data[cur]; data[cur] = 0; --cur; int child = 0; int i = 0; for(i = 1; i*2 <= cur; i = child) { child = i*2; if(child != cur && data[child+1] > data[child]) { ++child; } if(lastElement < data[child]) data[i] = data[child]; else break; } data[i] = lastElement; } int front()const { return data[1]; } bool isFull()const { return cur >= size; } bool isEmpty() { return cur == 0; }
};template <typename _Type>class MinHeap{private: _Type data[20]; int size; int cur;public: MinHeap(int _n):size(_n),cur(0) { memset(data,0,sizeof(data)); data[0] = -INF; } ~MinHeap() { } void clear() { memset(data,0,sizeof(data)); data[0] = -INF; size = N; cur = 0; } void push(_Type _value) { if(isFull()) return; int i ; cur++; for(i = cur; data[i/2] > _value; i/= 2) { data[i] = data[i/2]; } data[i] = _value; } void pop() { if(isEmpty()) return ; int child = 0; int i = 0; int lastElement = data[cur]; data[cur] = 0; --cur; for(i = 1; i*2 <= cur; i = child) { child = i*2; if(child != cur && data[child+1] < data[child]) { child = child+1; } if(lastElement > data[child]) data[i] = data[child]; else break; } data[i] = lastElement; } bool isFull()const { return cur >= size; } bool isEmpty()const { return cur == 0; } int front()const { return data[1]; }};int n,n1,n2;void Test(){ MinHeap<int> minHeap(n1); MaxHeap<int> maxHeap(n2); double sum = 0; int tmp; for(int i = 0; i < n; ++i) { scanf("%d",&tmp); if(minHeap.isFull()) { if(minHeap.front() < tmp) { minHeap.pop(); minHeap.push(tmp); } } else { minHeap.push(tmp); } if(maxHeap.isFull()) { if(maxHeap.front() > tmp) { maxHeap.pop(); maxHeap.push(tmp); } } else { maxHeap.push(tmp); } sum += tmp; } while(!maxHeap.isEmpty()) { sum -= maxHeap.front(); maxHeap.pop(); } while(!minHeap.isEmpty()) { sum -= minHeap.front(); minHeap.pop(); } printf("%.6lf\n",sum/(n-n1-n2));}int main(){ while(scanf("%d %d %d",&n1,&n2,&n) != EOF) { if(n1 == 0 && n2 == 0 && n == 0) break; Test(); } return 0;}
题目2： http://poj.org/problem?id=3125
题意：打印机取当前打印任务，若不是最高的打印任务，插到队尾，计算给定的任务在什么时候完成打印
题解：使用简单队列模拟，que[m] == 0表示完成了打印
代码：
#include <stdio.h>#include <algorithm>using namespace std;int que[200];void Test(){ int n,m; scanf("%d %d",&n,&m); for(int i = 0; i < n; ++i) { scanf("%d",&(que[i])); } int time = 0; int s = 0; while(que[m] > 0) { if(que[s] > 0) { int maxP = -1; for(int i = 0; i < n; ++i) { if(maxP < que[i]) { maxP = que[i]; } } if(que[s] >= maxP) { ++time; que[s] = 0; } } s = (s+1)%n; } printf("%d\n",time);}int main(){ //freopen("data.txt","r",stdin); int testcase; scanf("%d",&testcase); for(int i = 0; i < testcase; ++i) { Test(); } return 0;}
题目3： http://poj.org/problem?id=2318
题意：一个矩形，被n条直线划分，然后给定点(x,y)，判断所在区域
题解：点与直线的关系，二分法
代码：
#include <stdio.h>#include <string.h>const int N = 5005;int n,m,x1,y1,x2,y2;int U[N];int L[N];int cnts[N];void BinarySlove(int _x,int _y){ int low = -1; int high = n; int mid = 0; while(low + 1< high) { mid = low + (high - low)/ 2; double d = (double)(y1-y2)*(_x - U[mid]) + (double)(y1-_y)*(U[mid]-L[mid]); if(d > 0) { low = mid ; } else { high = mid ; } } cnts[high]++;}void Test(){ memset(cnts,0,sizeof(cnts)); for(int i = 0; i < n; ++i) { scanf("%d %d",&U[i],&L[i]); } int xT,yT; for(int i = 0; i < m; ++i) { scanf("%d %d",&xT,&yT); BinarySlove(xT,yT); } for(int i = 0; i <= n; ++i) { printf("%d: %d\n",i,cnts[i]); } printf("\n");}int main(){ //freopen("data.txt","r",stdin); while(scanf("%d ",&n) != EOF) { if(n == 0) break; scanf("%d %d %d %d %d",&m,&x1,&y1,&x2,&y2); Test(); } return 0;}
题目4： http://poj.org/problem?id=2106
题意：布尔表达式求解
题解：使用递归下降法LL求解
语法如下：
exp ---> (alt '|' )*alt
alt ---> (unit '&')*unit
unit ---> '('exp')' | '!'unit | 'F'|'V'
代码：
#include <stdio.h>#include <string.h>char context[256];int cur;int len;bool exp(char* _context);bool unit(char* _context){ if(_context[cur] == '!') { ++cur; return !unit(_context); } else if(_context[cur] == '(') { ++cur; bool ret = exp(_context); ++cur; return ret; } else if(_context[cur] == 'F') { ++cur; return false; } else { ++cur; return true; }}bool alt(char* _context){ bool ret = unit(_context); while(cur < len && _context[cur] == '&') { ++cur; bool ret2 = unit(_context); ret = ret && ret2; } return ret;}bool exp(char* _context){ bool ret = alt(_context); while(cur < len && _context[cur] == '|') { ++cur; bool ret2 = alt(_context); ret = ret || ret2; } return ret;}void Test(){ len = strlen(context); cur = 0; bool ret = exp(context); if(ret) printf("V\n"); else printf("F\n");}int main(){ //freopen("data.txt","r",stdin); char token; int testcase = 0; while(token = getchar(),token != EOF) { ++testcase; memset(context,0,sizeof(context)); int k = 0; while(token != '\n' && token != EOF) { if(token !=' ') { context[k++] = token; } token = getchar(); } if(strlen(context) > 0) { printf("Expression %d: ",testcase); Test(); } } return 0;}