题面

Garden
Time Limit: 1000MS Memory Limit: 65536K

Description
There are n flowerpots in Daniel's garden. These flowerpots are placed in n positions, and these n positions are numbered from 1 to n. Each flower is assigned an aesthetic value. The aesthetic values vary during different time of a day and different seasons of a year. Naughty Daniel is a happy and hardworking gardener who takes pleasure in exchanging the position of the flowerpots.
Friends of Daniel are great fans of his miniature gardens. Before they visit Daniel's home, they will take their old-fashioned cameras which are unable to adjust the focus so that it can give a shot of exactly k consecutive flowerpots. Daniel hopes his friends enjoy themselves, but he doesn't want his friend to see all of his flowers due to some secret reasons, so he guides his friends to the best place to catch the most beautiful view in the interval [x, y], that is to say, to maximize the sum of the aesthetics values of the k flowerpots taken in one camera shot when they are only allow to see the flowerpots between position x to position y.
There are m operations or queries are given in form of (p, x, y), here p = 0, 1 or 2. The meanings of different value of p are shown below.

1. p=0 settheaestheticvalueofthepotinpositionxasy.(1<=x<=n;-100<= y <= 100)
2. p = 1 exchange the pot in position x and the pot in position y. (1 <= x, y <= n; x might equal to y)
3. p = 2 print the maximum sum of aesthetics values of one camera shot in interval [x, y]. (1 <= x <= y <= n; We guarantee that y-x+1>=k)

Input
There are multiple test cases.
The first line of the input file contains only one integer indicates the number of test cases.
For each test case, the first line contains three integers: n, m, k (1 <= k <= n <= 200,000; 0 <= m <= 200,000).
The second line contains n integers indicates the initial aesthetic values of flowers from position 1 to position n. Some flowers are sick, so their aesthetic values are negative integers. The aesthetic values range from -100 to 100. (Notice: The number of position is assigned 1 to n from left to right.)
In the next m lines, each line contains a triple (p, x, y). The meaning of triples is mentioned above.

Output
For each query with p = 2, print the maximum sum of the aesthetics values in oneshot in interval [x, y].

Sample Input
1
5 7 3
-1 2 -4 6 1
2 1 5
2 1 3
1 2 1
2 1 5
2 1 4
0 2 4
2 1 5

Sample Output
4
-3
3
1
6

解题思路

本题难点主要在于 p = 2 的操作，即在区间 [x, y] 中查询长度为 k 的子区间内元素和的最大值，初看可能没有头绪，实际上我们可以把它转化成基本的区间查询。方法是：用 sum[i] 表示从 i 开始的长度为 k 的区间内的元素和，将原先一段固定长度的区间和转化成一个元素的值。这样我们就可以使用线段树来进行区间最值查询了。

以示例输入为例，我们将原数组转化成 sum[1] = -3, sum[2] = 4, sum[3] = 3 。这样当我们进行操作 2 1 5 时，实际上只要查询 sum[1] ~ sum[3] 中的最大值。

解决了如何实现操作 2 的问题，我们再来看一下应该如何实现操作 0 ，即更新下标为 x 处的值为 y。不妨思考一下，将其代入到刚才转化的模型中看，修改单个元素的值会对 sum 数组造成怎样的影响呢？不难发现，修改操作会影响到所有包含此元素的区间，这些区间的元素和会同时变化，变化值为被修改元素的变化值 d 。这样一来就修改操作就很容易实现了，只需要计算出被修改元素的变化值 d = y - a[x] ，然后将所有包含此元素的 sum[i] 都使用线段树更新，加上变化值 d 即可。

对于操作 1，等价于两次操作 0 ，在此不再赘述。

参考代码

#include <cstdio>
#include <algorithm>
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
const int MAXN = 200001;
const int INF = 0x3f3f3f3f;
using namespace std;

int lazy[MAXN<<2];
int MAX[MAXN<<2];
int a[MAXN], sum[MAXN];

void PushUp(int rt) {
    MAX[rt] = max(MAX[rt<<1], MAX[rt<<1|1]);
}

void PushDown(int rt) {
    if(lazy[rt]) {
        lazy[rt<<1] += lazy[rt];
        lazy[rt<<1|1] += lazy[rt];
        MAX[rt<<1] += lazy[rt];
        MAX[rt<<1|1] += lazy[rt];
        lazy[rt] = 0;
    }
}

void Build(int l, int r, int rt) {
    lazy[rt] = 0;
    if (l == r) {
        MAX[rt] = sum[l];
        return;
    }
    int m = (l + r) >> 1;
    Build(lson);
    Build(rson);
    PushUp(rt);
}

void Update(int L, int R, int v, int l, int r, int rt) {
    if(L<=l && r<=R) {
        lazy[rt] += v;
        MAX[rt] += v;
        return;
    }
    PushDown(rt);
    int m = (l + r) >> 1;
    if(L <= m) Update(L, R, v, lson);
    if(m < R) Update(L, R, v, rson);
    PushUp(rt);
}

int QueryMAX(int L, int R, int l, int r, int rt) {
    if(L <= l && r <= R) {
        return MAX[rt];
    }
    PushDown(rt);
    int m = (l + r) >> 1;
    int ret = -INF;
    if(L <= m) ret = max(ret, QueryMAX(L, R, lson));
    if(m < R) ret = max(ret, QueryMAX(L, R, rson));
    return ret;
}

int main(int argc, char const *argv[]) {
    int t, n, m, k, p, x, y;
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d %d", &n, &m, &k);
        for(int i=1; i<=n; ++i) {
            scanf("%d", &a[i]);
        }
        // 计算区间内元素和数组 sum
        sum[1] = 0;
        for(int i=1; i<=k; ++i) {
            sum[1] += a[i];
        }
        for(int i=2; i<=n-k+1; ++i) {
            sum[i] = sum[i-1]-a[i-1] + a[i+k-1];
        }
        // 重设 n 为 sum 数组的长度
        n = n-k+1;
        // 构造线段树
        Build(1, n, 1);
        while(m--) {
            scanf("%d %d %d", &p, &x, &y);
            if(p == 0) {
                int d = y-a[x];
                a[x] = y;
                Update(max(1, x-k+1), x, d, 1, n, 1);
            }
            if(p == 1) { // 两次操作 0
                int tmp = a[x];
                int d = a[y]-a[x];
                a[x] = a[y];
                Update(max(1, x-k+1), x, d, 1, n, 1);
                d = tmp-a[y];
                a[y] = tmp;
                Update(max(1, y-k+1), y, d, 1, n, 1);
            }
            if(p == 2)
                printf("%d\n", QueryMAX(x, y-k+1, 1, n, 1));
        }
    }
    
    return 0;
}