分治法-求最大最小元

void maxmin(int A[], int &e_max, int &e_min ,int low, int high)
{
int mid, x1, y1, x2, y2;
if (high - low <= 1) {　　　　　　　　　　　　　//<=,可能为1个
if (A[high] > A[low]) {
e_max = A[high];
e_min = A[low];
}
else {
e_max = A[low];
e_min = A[high];
}
}                                       　　  //特征1：小到容易求解
　　else { 　　　　　　　　　　　　　　　　　　　　
　　　　 mid = (low + high) / 2;　　　　　　　　　　　//特征2：可分解　
　　　　 maxmin(A, x1, y1, low, mid);              //特征4：子问题独立
　　　　 maxmin(A, x2, y2, mid + 1, high);
　　　　e_max = max(x1, x2);                      //特征3：可合并
　　　　 e_min = min(y1, y2);
　　 }
}

#include <stdio.h>
#include<stdlib.h>

int max(int a, int b)
{
return (a > b) ? a : b;
}

int min(int a, int b)
{
return (a < b) ? a : b;
}

void maxmin(int A[], int &e_max, int &e_min ,int low, int high)
{
int mid, x1, y1, x2, y2;
if (high - low <= 1) {                //<=,可能为1个
if (A[high] > A[low]) {
e_max = A[high];
e_min = A[low];
}
else {
e_max = A[low];
e_min = A[high];
}
}                                                //特征1：小到容易求解
else {
mid= (low + high) / 2;            //特征2：可分解　
maxmin(A, x1, y1, low, mid);    //特征4：子问题独立
maxmin(A, x2, y2, mid + 1, high);
e_max = max(x1, x2);            //特征3：可合并
e_min =  min(y1, y2);
}
}

int main()
{
int init[11], max, min;
for(int i = 1; i <= 10; i ++){
scanf("%d", &init[i]);
}
maxmin(init, max, min, 1, 10);
printf("%d %d", max, min);
system("pause");
}

/*
(0 1 3 5 2 4 6 8 9 7),ENTER LEVEL ONE
1          5                    ENTER LEVEL TWO
1    3                        ENTER LEVEL THREE
1 2                            ENTER LEVEL FOUR
e_max = 1;
e_min = 0;
1 2                            EXIT LEVEL FOUR            x1 = 1; y1 = 0;

3
e_max = e_min =3;
3                        EXIT LEVEL FOUR            x2 = y2 = 3;

e_max = 3;  -> x1;
e_min = 0;   -> y1;
1    3                        EXIT LEVEL THREE            x1 = 1; y1 = 3;

4 5                    ENTER  LEVEL THREE
x2 = 5;
y2 = 2;
e_max = 5;  -> x1;
e_min = 0;   -> y1;
4 5                     EXIT LEVEL THREE        x2 = 5; y2 = 2;

e_max = 5;
e_min = 0;
1           5                  EXIT LEVEL TWO            x1 = 5; y1 = 0;

6          10 ENTER LEVEL TWO
6    8         ENTER LEVEL THREE
6 7            ENTER LEVEL FOUR
e_max = 6;
e_min = 4;
6 7            EXIT LEVEL FOUR            x1 = 6; y1 = 4;
8         ENTER LEVEL FOUR
e_max = e_min = 8;
8         EXIT LEVEL FOUR            x2 = y2 = 8;
e_max = 8;
e_min = 4;
6    8         EXIT LEVEL THREE        x1 = 8; y1 = 4;
9 10 ENTER LEVEL THREE
e_max = 9;
e_min = 7;
9 10 EXIT LEVEL THREE        x2 = 9; y2 = 7;
6          10 EXIT LEVEL TWO            x2 = 9; y2 = 4;

e_max = 9;
e_min = 0;
1                           10  EXIT LEVEL ONE        e_max = 9; e_min = 0;

*/