一道序列算法题

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题目：

有a和b两个严格增序列，a和b元素相同称为交汇点。例子中 5,9,22为交汇点，在交汇点处可以切换路径。

从a或b首元素出发，求元素之和最大的一条路径。例子中1-3-5-7-8-9-18-19-22-28-40为元素和最大的一条路径

a[] = {1,3,5,9,20,22,28,40};b[] = {5,7,8,9,18,19,22,27};

//题目：有a和b两个严格增序列，a和b元素相同称为交汇点。//例子中 5,9,22为交汇点，在交汇点处可以切换路径。//从a或b首元素出发，求元素之和最大的一条路径。//例子中1-3-5-7-8-9-18-19-22-28-40为元素和最大的一条路径//a[] = {1,3,5,9,20,22,28,40};b[] = {5,7,8,9,18,19,22,27};#include <stdio.h>struct intersection{int *a;int *b;};#define MAXLEN  50int sum_array(struct intersection x, struct intersection x_next);int array_max(int *a, int m, int *b ,int n){int i;int *pa = a; int *pb = b;int k = 1;int sum = 0; struct intersection intersec[MAXLEN];   //存放交汇点指针intersec[0].a = a - 1;      //设第一个交汇点在首元素之前intersec[0].b = b - 1;for(i = 0; i < m; i++){while((*pa > *pb) && (pb < b + n -1))pb++;if(*pa == *pb){intersec[k].a = pa;intersec[k].b = pb;k++; }pa++;if(pb == b + n - 1)      //end of array bbreak;}intersec[k].a = a + m;    //设最后一个交汇点在末元素之后intersec[k].b = b + n;for(i = 0; i < k; i++){sum += sum_array(intersec[i], intersec[i+1]);  //a和b在两交汇点间的元素（不包含交汇点）分别求和，并取最大值}for(i = 1; i < k; i++){sum += *(intersec[i].a); printf("intersection:%d\n", *(intersec[i].a));}return sum;}int sum_array(struct intersection x, struct intersection x_next){int *p;int max;int sum_A = 0; int sum_B = 0;for(p = x.a + 1; p <=x_next.a - 1; p++)sum_A += *p;for(p = x.b + 1; p <=x_next.b - 1; p++)sum_B += *p;max = sum_A>sum_B? sum_A : sum_B;return max;}int main(void){int ret, i;int m, n;int a[] = {1,3,5,9,20,22,28,40,80,90,99};int b[] = {5,7,8,9,18,19,22,27,41,43,90,96,98};m =sizeof(a)/sizeof(int); n =sizeof(b)/sizeof(int);for (i = 0; i < m; ++i){if(i == 0)   printf("a[] = %d", a[i]);elseprintf("-%d", a[i]);}printf("\n");for (i = 0; i < n; ++i){if(i == 0)   printf("b[] = %d", b[i]);elseprintf("-%d", b[i]);}printf("\n");ret = array_max(a, m, b, n);printf("\nmax = %d\n", ret);return 0;}

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