BNUOJ39566 Do use segment tree （树链剖分+维护区间最大连续和）

Do use segment tree

Given a tree with n (1 ≤ n ≤ 200,000) nodes and a list of q (1 ≤ q ≤ 100,000) queries, process the queries in order and output a value for each output query. The given tree is connected and each node on the tree has a weight wi (-10,000 ≤ wi ≤ 10,000).

Each query consists of a number ti (ti = 1, 2), which indicates the type of the query , and three numbers ai, bi and ci (1 ≤ ai, bi ≤ n, -10,000 ≤ ci ≤ 10,000). Depending on the query type, process one of the followings:

(ti = 1: modification query) Change the weights of all nodes on the shortest path between ai and bi (both inclusive) to ci.

(ti = 2: output query) First, create a list of weights on the shortest path between ai and bi (both inclusive) in order. After that, output the maximum sum of a non-empty continuous subsequence of the weights on the list. ci is ignored for output queries.

Input

The first line contains two integers n and q. On the second line, there are n integers which indicate w1, w2, … , wn.

Each of the following n - 1 lines consists of two integers si and ei (1 ≤ si, ei ≤ n), which means that there is an edge between si and ei.

Finally the following q lines give the list of queries, each of which contains four integers in the format described above. Queries must be processed one by one from top to bottom.

Output

For each output query, output the maximum sum in one line.

Sample Input 1

3 4
1 2 3
1 2
2 3
2 1 3 0
1 2 2 -4
2 1 3 0
2 2 2 0

Output for the Sample Input 1

6
3
-4

Sample Input 2

7 5
-8 5 5 5 5 5 5
1 2
2 3
1 4
4 5
1 6
6 7
2 3 7 0
2 5 2 0
2 4 3 0
1 1 1 -1
2 3 7 0

Output for the Sample Input 2

12
10
10
19

Sample Input 3

21 30
10 0 -10 -8 5 -5 -4 -3 1 -2 8 -1 -7 2 7 6 -9 -6 3 4 9
10 3
3 2
3 12
12 4
4 13
4 9
10 21
21 1
1 11
11 14
1 15
10 6
6 17
6 16
6 5
5 18
5 19
10 7
10 8
8 20
1 1 21 -10
1 3 19 10
2 1 13 0
1 4 18 8
1 5 17 -5
2 16 7 0
1 6 16 5
1 7 15 4
2 4 20 0
1 8 14 3
1 9 13 -1
2 9 18 0
1 10 12 2
1 11 11 -8
2 21 15 0
1 12 10 1
1 13 9 7
2 6 14 0
1 14 8 -2
1 15 7 -7
2 10 2 0
1 16 6 -6
1 17 5 9
2 12 17 0
1 18 4 6
1 19 3 -3
2 11 8 0
1 20 2 -4
1 21 1 -9
2 5 19 0

Output for the Sample Input 3

20
9
29
27
10
12
1
18
-2
-3

• 树链剖分转化为线段树；

• 线段树上维护最大前缀和，最大后缀和，最大区间连续和，区间和。

  剩下部分就随便搞搞。

• 代码 36 ms 31040 KB 4521 B 2015-10-11 00:45:56

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int inf = 0x3f3f3f3f;const int maxn = 200000 + 10;//treeint head[maxn], to[maxn<<1], nxt[maxn<<1], tot = 0, sz = 0;void addedge(int _u, int _v){    to[++tot] = _v; nxt[tot] = head[_u]; head[_u] = tot;}//divide treeint siz[maxn], fa[maxn], dep[maxn];void dfs1(int u){    siz[u] = 1;    for(int i = head[u]; i; i = nxt[i]){        int v = to[i];        if (v == fa[u]) continue;        dep[v] = dep[u] + 1;        fa[v] = u;        dfs1(v);        siz[u] += siz[v];    }}int pos[maxn], belong[maxn], szu[maxn];void dfs2(int u, int f){    int k = 0;    pos[u] = ++sz; szu[sz] = u; belong[u]= f;    for(int i = head[u]; i; i = nxt[i]){        int v = to[i];        if (dep[v] > dep[u] && siz[v] > siz[k]) k = v;    }    if (!k) return;    dfs2(k, f);    for(int i = head[u]; i; i = nxt[i]){        int v = to[i];        if (dep[v] > dep[u] && v != k) dfs2(v, v);    }}#define Lson (k<<1)#define Rson ((k<<1)|1)//segment treestruct Seg{    int l, r, prefix, suffix, maxseq, sum, setv;    void setVal(int val){        int tmp = val * (r - l + 1);        sum = tmp;        prefix = suffix = maxseq = max(tmp, val);        setv = val;    }}t[maxn<<2];Seg operator + (Seg a, Seg b){    if (a.maxseq == -inf) return b;    if (b.maxseq == -inf) return a;    Seg re;    re.sum = a.sum + b.sum;    re.prefix = max(a.prefix, a.sum + b.prefix);    re.suffix = max(b.suffix, a.suffix + b.sum);    re.maxseq = max(max(a.maxseq, b.maxseq), a.suffix + b.prefix);    return re;}int w[maxn];void pushup(int k){    t[k].sum = t[Lson].sum + t[Rson].sum;    t[k].prefix = max(t[Lson].prefix, t[Lson].sum + t[Rson].prefix);    t[k].suffix = max(t[Rson].suffix, t[Lson].suffix + t[Rson].sum);    t[k].maxseq = max(max(t[Lson].maxseq, t[Rson].maxseq), t[Lson].suffix + t[Rson].prefix);}void build(int k, int l, int r){    t[k].setv = inf;    t[k].l = l; t[k].r = r;    if (l==r){        t[k].prefix = t[k].suffix = t[k].maxseq = t[k].sum = w[szu[l]];        return;    }    int mid = (l+r)>>1;    build(Lson, l, mid);    build(Rson, mid+1, r);    pushup(k);}void pushdown(int k){    if (t[k].setv != inf){        t[Lson].setVal(t[k].setv);        t[Rson].setVal(t[k].setv);        t[k].setv = inf;    }}void change(int k, int x, int y, int val){    int l = t[k].l, r = t[k].r, mid = (l+r)>>1;    if (x <= l && r <= y){        t[k].setVal(val);        return;    }    pushdown(k);    if (y <= mid) change(Lson, x, y, val);    else if (x > mid) change(Rson, x, y, val);    else change(Lson, x, mid, val), change(Rson, mid+1, y, val);    pushup(k);}void solveChange(int x, int y, int k){    while(belong[x] != belong[y]){        if (dep[belong[x]] < dep[belong[y]]) swap(x, y);        change(1, pos[belong[x]], pos[x], k);        x = fa[belong[x]];    }    if (dep[x] < dep[y]) swap(x, y);    change(1, pos[y], pos[x], k);}Seg query(int k, int x, int y){    int l = t[k].l, r = t[k].r, mid = (l+r)>>1;    if (x <= l && r <= y){        return t[k];    }    pushdown(k);    if (y <= mid) return query(Lson, x, y);    else if (x > mid) return query(Rson, x, y);    else return query(Lson, x, mid) + query(Rson, mid+1, y);}void solveMax(int x, int y){    Seg tx, ty;    tx.maxseq = -inf;    ty.maxseq = -inf;    while(belong[x] != belong[y]){        if (dep[belong[x]] > dep[belong[y]]){            tx = query(1, pos[belong[x]], pos[x]) + tx;            x = fa[belong[x]];        }        else{            ty = query(1, pos[belong[y]], pos[y]) + ty;            y = fa[belong[y]];        }    }    if (dep[x] > dep[y]){        tx = query(1, pos[y], pos[x]) + tx;    }    else{        ty = query(1, pos[x], pos[y]) + ty;    }    swap(tx.prefix, tx.suffix);    tx = tx + ty;    printf("%d\n", tx.maxseq);}int main(){    int n, Q, op, x, y, z;    scanf("%d%d", &n, &Q);    for(int i = 1; i <= n; i++){        scanf("%d", &w[i]);    }    for(int _i = 1; _i < n; _i++){        scanf("%d%d", &x, &y);        addedge(x, y);        addedge(y, x);    }    dfs1((1+n)/2);    dfs2((1+n)/2, (1+n)/2);    build(1, 1, n);    while(Q--){        scanf("%d%d%d%d", &op, &x, &y, &z);        if (op == 1){            solveChange(x, y, z);        }        else{            solveMax(x, y);        }    }    return 0;}