传递参数时引发的异常

时间:2019-01-27 08:12:18

标签: c++ segment-tree

我遇到错误Exception thrown at 0x009523B9 in Project5.exe: 0xC00000FD: Stack overflow (parameters: 0x00000001, 0x00E42ED4). 在下面的程序中

测试案例:

  

5

     

1 2 3 4 5

     

1

     

Q 2 4

执行此行时出现错误 struct node query(int index, int start, int endv, int l, int r) 这是在main中第一次调用query函数的时候

该参数就像抛出异常时一样 query(0,0,4,1,3)

,但它们会自动更改为query(0,1,1,1,3) 我在哪里做错了

这是基于段树spoj KGSS

的问题
#include<iostream>
#include<algorithm>
#include<climits>
#pragma warning(disable:4996)    //disable scanf warning 
using namespace std;

struct node {
    long long int maxsum;
    long long int maxnum;
};
long long int a[100];
struct node tree[100];
void build(int index, int start, int endv) {
    if (start == endv) {
        tree[index].maxsum = a[start];
        tree[index].maxnum = a[start];
    }
    else {
        int mid = (start + endv) / 2;
        build(2 * index + 1, start, mid);
        build(2 * index + 2, mid + 1, endv);

        tree[index].maxnum = max(tree[2 * index + 1].maxnum, tree[2 * index + 2].maxnum);
        tree[index].maxsum = max(tree[2 * index + 1].maxnum + tree[2 * index + 2].maxnum, max(tree[2 * index + 1].maxsum, tree[2 * index + 2].maxsum));
    }
}

struct node query(int index, int start, int endv, int l, int r) {

    struct node result;

    result.maxnum = INT_MIN;
    result.maxsum = INT_MIN;

    if (l > endv || r < start)
        return result;
    int mid = (start + endv) / 2;
    struct node left, right;
    left = query(index, start, mid, l, r);
    right = query(index, mid + 1, endv, l, r);

    tree[index].maxnum = max(left.maxnum, right.maxnum);
    tree[index].maxsum = max(left.maxnum + right.maxnum, max(left.maxsum, right.maxsum));

    return tree[index];
}

void update(int index, int start, int endv, int aidx, long long int value) {


    if (start == endv) {
        a[aidx] = value;
        tree[index].maxsum = value;
        tree[index].maxnum = value;
    }
    else {

        int mid = (start + endv) / 2;
        if (aidx <= mid) {
            update(2 * index + 1, start, mid, aidx, value);
        }
        else {
            update(2 * index + 2, mid + 1, endv, aidx, value);
        }

        tree[index].maxnum = max(tree[2 * index + 1].maxnum, tree[2 * index + 2].maxnum);
        tree[index].maxsum = max(tree[2 * index + 1].maxnum + tree[2 * index + 2].maxnum, max(tree[2 * index + 1].maxsum, tree[2 * index + 2].maxsum));
    }
}

int main() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
        scanf("%d", &a[i]);
    }
    build(0, 0, n - 1);
    //for(int i = 0)
    int q;
    cin >> q;
    for (int i = 0; i < q; i++) {
        int l, r;
        char ap;
        cin >> ap >> l >> r;
        if (ap == 'U') {
            update(0, 0, n - 1, l - 1, r);
        }
        else {
            struct node temp = query(0, 0, n - 1, l - 1, r - 1);
            printf("%lld\n", temp.maxsum);
        }
    }

}

注意->我使用了result.maxnum = INT_MIN; result.maxsum = INT_MIN;

在查询函数中,result.maxnumresult.maxsum属于long long int类型,我正在为其分配INT_MIN

这是因为有时候从递归的两边我都会得到最小值 将这些值相加并将它们存储在int变量中会导致整数溢出,这就是为什么我使用long long int可以有人提供处理这种情况的方式还是可以的

1 个答案:

答案 0 :(得分:0)

    scanf("%d", &a[i]);
a[i]的类型为long long int时,

不正确。将C ++ I / O与C I / o混合使用也不是一个好主意。没什么问题

cin >> a[i];

但是主要问题是您有堆栈溢出。在您的示例中

query(0,0,4,1,3) ->
 query(0,0,2,1,3) ->
  query(0,0,1,1,3) ->
   query(0,0,0,1,3) ->
    return result
   query(0,1,1,1,3) ->
    query(0,1,1,1,3) ->
     query(0,1,1,1,3) ->
      query(0,1,1,1,3) ->
       query(0,1,1,1,3) ->
        query(0,1,1,1,3) ->
         query(0,1,1,1,3) ->
          BOOM!! stack overflow

由于我不了解代码,因此不太确定如何解决此问题。我本来会以为

if (start >= endv)
    return result;

一开始会有所帮助,但我不确定。