[bzoj3531][Sdoi2014]旅行

题目大意:给你一棵树,每个点有一个分类和一个值。有四种操作:

  1. 修改某个点的分类
  2. 修改某个点的值
  3. 查询两个分类相同的点的最短路上,与这两个点分类相同的所有点的值的和
  4. 查询两个分类相同的点的最短路上,与这两个点分类相同的所有点的值的最大值

题解:树链剖分,对每一个分类建一棵动态开点线段树就好了。

卡点:1.询问是传根写成了传编号

 

C++ Code:

#include <cstdio>
#define maxn 100010
#define N 100010 * 4 * 20
using namespace std;
int n, Q;
int w[maxn], c[maxn];
inline int max(int a, int b) {return a > b ? a : b;}
inline void swap(int &a, int &b) {a ^= b ^= a ^= b;}

int head[maxn], cnt;
struct Edge {
    int to, nxt;
} e[maxn << 1];
void add(int a, int b) {
    e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
}

int sz[maxn], fa[maxn], dep[maxn], son[maxn];
int top[maxn], dfn[maxn], idx;
void dfs1(int rt) {
    sz[rt] = 1;
    for (int i = head[rt]; i; i = e[i].nxt) {
        int v = e[i].to;
        if (!dep[v]) {
            dep[v] = dep[rt] + 1;
            fa[v] = rt;
            dfs1(v);
            sz[rt] += sz[v];
            if (!son[rt] || sz[v] > sz[son[rt]]) son[rt] = v;
        } 
    }
}
void dfs2(int rt) {
    dfn[rt] = ++idx;
    int v = son[rt];
    if (v) top[v] = top[rt], dfs2(v);
    for (int i = head[rt]; i; i = e[i].nxt) {
        v = e[i].to;
        if (v != fa[rt] && v != son[rt]) {
            top[v] = v;
            dfs2(v);
        }
    }
}

int root[maxn], M[N], S[N];
int rc[N], lc[N], tot;
void add(int &rt, int l, int r, int p, int num) {
    if (!rt) rt = ++tot;
    if (l == r) {
        S[rt] = M[rt] = num;
        return ;
    }
    int mid = l + r >> 1;
    if (p <= mid) add(lc[rt], l, mid, p, num);
    else add(rc[rt], mid + 1, r, p, num);
    M[rt] = max(M[lc[rt]], M[rc[rt]]);
    S[rt] = S[lc[rt]] + S[rc[rt]];
}
int askM(int rt, int l, int r, int L, int R) {
    if (!rt || l > r || L > R) return 0;
    if (L <= l && R >= r) return M[rt];
    int mid = l + r >> 1, ans = 0;
    if (L <= mid) ans = askM(lc[rt], l, mid, L, R);
    if (R > mid) ans = max(ans, askM(rc[rt], mid + 1, r, L, R));
    return ans;
}
int askS(int rt, int l, int r, int L, int R) {
    if (!rt || l > r || L > R) return 0;
    if (L <= l && R >= r) return S[rt];
    int mid = l + r >> 1, ans = 0;
    if (L <= mid) ans = askS(lc[rt], l, mid, L, R);
    if (R > mid) ans += askS(rc[rt], mid + 1, r, L, R);
    return ans;
}

int queryM(int rt, int x, int y) {
    int ans = 0;
    while (top[x] != top[y]) {
        if (dep[top[x]] < dep[top[y]]) swap(x, y);
        ans = max(ans, askM(rt, 1, n, dfn[top[x]], dfn[x]));
        x = fa[top[x]];
    }
    if (dep[x] > dep[y]) swap(x, y);
    ans = max(ans, askM(rt, 1, n, dfn[x], dfn[y]));
    return ans;
}
int queryS(int rt, int x, int y) {
    int ans = 0;
    while (top[x] != top[y]) {
        if (dep[top[x]] < dep[top[y]]) swap(x, y);
        ans += askS(rt, 1, n, dfn[top[x]], dfn[x]);
        x = fa[top[x]];
    }
    if (dep[x] > dep[y]) swap(x, y);
    ans += askS(rt, 1, n, dfn[x], dfn[y]);
    return ans;
}

int main() {
    scanf("%d%d", &n, &Q);
    for (int i = 1; i <= n; i++) scanf("%d%d", &w[i], &c[i]);
    for (int i = 1; i < n; i++) {
        int a, b;
        scanf("%d%d", &a, &b);
        add(a, b);
        add(b, a);
    }
    dep[top[1] = 1] = 1;
    dfs1(1);
    dfs2(1);
    for (int i = 1; i <= n; i++) add(root[c[i]], 1, n, dfn[i], w[i]);
    while (Q --> 0) {
        char op[10];
        int x, y;
        scanf("%s%d%d", op, &x, &y);
        if (op[1] == 'C') {
            add(root[c[x]], 1, n, dfn[x], 0);
            c[x] = y;
            add(root[y], 1, n, dfn[x], w[x]);
        }
        if (op[1] == 'W') {
            w[x] = y;
            add(root[c[x]], 1, n, dfn[x], y);
        }
        if (op[1] == 'S') {
            printf("%d\n", queryS(root[c[x]], x, y));
        }
        if (op[1] == 'M') {
            printf("%d\n", queryM(root[c[x]], x, y));
        }
    }
    return 0;
}

  

posted @ 2018-08-17 15:18  Memory_of_winter  阅读(114)  评论(0编辑  收藏  举报