HDU1166 敌兵布阵（单点更新 区间查询）

敌兵布阵

这道题就是单点更新，区间查询的裸题，比较入门。告诉你一段区间，然后查询[l,r]的最值，区间内的值是允许修改的。

线段树做法

#include
#define maxn 50005
struct node
{
    int l, r, sum;
} tr[4 * maxn];
int arr[maxn];
void build(int d, int l,int r)
{
    tr[d].l = l;
    tr[d].r = r;
    if(l == r)
    {
        tr[d].sum = arr[l];
        return ;
    }
    int mid = (l + r) / 2;
    int ld = 2 * d;
    int rd = 2 * d + 1;
    build(ld, l, mid);
    build(rd, mid + 1, r);
    tr[d].sum = tr[ld].sum + tr[rd].sum;
}
int query(int d, int l, int r)
{
    if(tr[d].l == l && tr[d].r == r)
    {
        return tr[d].sum;
    }
    int mid = (tr[d].l + tr[d].r) / 2;
    int ld = 2 * d;
    int rd = 2 * d + 1;
    if(r <= mid) return query(ld, l, r);
    else if(l > mid)return query(rd, l, r);
    else return query(ld, l, mid) + query(rd, mid + 1, r);
}
void update(int d, int val, int pos)
{
    if(tr[d].l == tr[d].r && tr[d].l == pos)
    {
        tr[d].sum += val;
        return ;
    }
    int mid = (tr[d].l + tr[d].r) / 2;
    int ld = 2 * d;
    int rd = 2 * d + 1;
    if(pos <= mid) update(ld, val, pos);
    else update(rd, val, pos);
    tr[d].sum = tr[ld].sum + tr[rd].sum;
}
int main()
{
    int t;
    while(~scanf("%d",&t))
    {
        for(int i = 1; i <= t; i++)
        {
            memset(arr, 0, sizeof(arr));
            memset(tr, 0, sizeof(tr));
            printf("Case %d:\n", i);
            int n;
            scanf("%d", &n);
            for(int j = 1; j <= n; j++)
            {
                scanf("%d", &arr[j]);
            }
            build(1, 1, n);
            char s[10];
            while(scanf("%s",s) && s[0] != 'E')
            {
                if(s[0] == 'Q')
                {
                    int x, y;
                    scanf("%d%d", &x, &y);
                    printf("%d\n", query(1, x, y));
                }
                else if(s[0] == 'A')
                {
                    int pos, val;
                    scanf("%d%d", &pos, &val);
                    update(1, val, pos);
                }
                else
                {
                    int pos, val;
                    scanf("%d%d", &pos, &val);
                    update(1, 0-val, pos);
                }
            }
        }
    }
}

树状数组做法：

#include
int c[50005];
int n;
int lowbit(int k)
{
    return k & (-k);
}
void update(int val, int k)
{
    while(k <= n)
    {
        c[k] += val;
        k += lowbit(k);
    }
}
int sum(int k)
{
    int res = 0;
    while(k > 0)
    {
        res += c[k];
        k -= lowbit(k);
    }
    return res;
}
int main()
{
    int t;
    scanf("%d", &t);
    for(int i = 1; i <= t; i++)
    {
        printf("Case %d:\n",i);
        memset(c, 0, sizeof(c));
        scanf("%d", &n);
        int temp;
        for(int j = 1; j <= n; j++)
        {
            scanf("%d", &temp);
            update(temp, j);
        }


    char s[6];
    while(scanf("%s",s) && s[0] != 'E')
    {
        if(s[0] == 'Q')
        {
            int x, y;
            scanf("%d%d", &x, &y);
            printf("%d\n", sum(y) - sum(x - 1));
        }
        else if(s[0] == 'A')
        {
            int x, y;
            scanf("%d%d", &x, &y);
            update(y, x);
        }
        else
        {
            int x, y;
            scanf("%d%d", &x, &y);
            update(-y, x);
        }
    }
    }
}