POJ 3237 Tree

学了一下树链剖分。就是把树剖成链,然后用线段树、树状数组、splay等数据结构来维护。

// POJ 3237 TREE


/**DESC: 给出一棵树,有三种操作:
      1:第i条边的权值修改成v.
      2:a 到 b 的路径上的权值全都取反。
      3:在 a 到 b的路径上的权值找最大。
*/

/** 思路:线段树维护树链剖分。
 * 道理我都懂,就是代码麻烦了有点。T_T
 */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <vector>
#define maxn 10010
using namespace std;

struct Edge{
    int u, v;
    int nxt;
}edge[maxn*2];

int head[maxn];
int tot;

void addEdge(int u, int v) {
    edge[tot].u = u;
    edge[tot].v = v;
    edge[tot].nxt = head[u];
    head[u] = tot++;
}

//树链剖分 把树剖成链
int fa[maxn]; //dfs1
int deep[maxn];
int num[maxn];
int son[maxn];

int top[maxn]; //dfs2
int p[maxn];
int fp[maxn];
int pos;

void dfs1(int u, int pre, int d) {
    deep[u] = d;
    fa[u] = pre;
    num[u] = 1;
    for (int i=head[u]; i!=-1; i=edge[i].nxt) {
        int v = edge[i].v;
        if (v != pre) {
            dfs1(v, u, d+1);
            num[u] += num[v];
            if (son[u] == -1 || num[v] > num[son[u]]) {
                son[u] = v;
            }
        }
    }
}

void dfs2(int u, int sp) {
    top[u] = sp;
    p[u] = pos++; /// u和父亲结点的边在线段树中的位置
    fp[p[u]] = u; /// 和fa[]数组相反,线段树中的第fp[u]条边是原树中u点和父亲的连边
    if (son[u] == -1) return;
    dfs2(son[u], sp);
    for (int i=head[u]; i!=-1; i=edge[i].nxt) {
        int v = edge[i].v;
        if (v != son[u] && v != fa[u]) {
            dfs2(v, v);
        }
    }
}

//线段树
struct Node {
    int l, r;
    int maxm;
    int minn;
    int ne;
}segTree[maxn*4];

void build(int rt, int l, int r) {
    segTree[rt].l = l;
    segTree[rt].r = r;
    segTree[rt].maxm = 0;
    segTree[rt].minn = 0;
    if (l == r) return;
    int mid = ((l+r)>>1);
    build(rt<<1, l, mid);
    build((rt<<1)|1, mid+1, r);
}

void push_down(int i) {
    if (segTree[i].l == segTree[i].r) return;
    if (segTree[i].ne) {
        segTree[i<<1].maxm = -segTree[i<<1].maxm;
        segTree[i<<1].minn = -segTree[i<<1].minn;
        swap(segTree[i<<1].maxm, segTree[i<<1].minn);
        segTree[(i<<1)|1].maxm = -segTree[(i<<1|1)].maxm;
        segTree[(i<<1)|1].minn = -segTree[(i<<1|1)].minn;
        swap(segTree[(i<<1)|1].maxm, segTree[(i<<1)|1].minn);

        segTree[i<<1].ne ^= 1; //左右子结点的延迟标记更新
        segTree[(i<<1)|1].ne ^= 1;
        segTree[i].ne = 0; ////
    }
}

void push_up(int i) {
    segTree[i].maxm = max(segTree[i<<1].maxm, segTree[(i<<1)|1].maxm);
    segTree[i].minn = min(segTree[i<<1].minn, segTree[(i<<1)|1].minn);
}

void update(int i, int k, int val) {
    if (segTree[i].l == k && segTree[i].r == k) {
        segTree[i].maxm = val;
        segTree[i].minn = val;
        segTree[i].ne = 0;///
        return;
    }
    push_down(i); //向下延迟标记
    int mid = (segTree[i].l + segTree[i].r) / 2;
    if (k <= mid) update(i<<1, k, val);
    else update((i<<1)|1, k, val);
    push_up(i); //向上延迟标记
}

void init() {
    memset(head, -1, sizeof(head));
    tot = 0;
    pos = 0;
    memset(son, -1, sizeof(son));
}

int e[maxn][3];

void ne_update(int rt, int l, int r) { //把线段树的[l, r]区间取反
    if (segTree[rt].l == l && segTree[rt].r == r) {
        segTree[rt].maxm = -segTree[rt].maxm;
        segTree[rt].minn = -segTree[rt].minn;
        swap(segTree[rt].maxm, segTree[rt].minn);
        segTree[rt].ne ^= 1; // 延迟标记
        return;
    }
    push_down(rt); ///
    int mid = (segTree[rt].l + segTree[rt].r) / 2;
    if (r <= mid) { //全都在左区间
        ne_update(rt<<1, l, r);
    }else if (l > mid) {
        ne_update((rt<<1)|1, l, r);
    }else {
        ne_update(rt<<1, l, mid);
        ne_update((rt<<1)|1, mid+1, r);
    }
    push_up(rt); ///
}


void Negate(int u, int v) {
    int f1 = top[u], f2 = top[v];
    while(f1 != f2) {
        if (deep[f1] < deep[f2]) { //使得depp[f1] > deep[f2]
            swap(f1, f2);
            swap(u, v);
        }
        ne_update(1, p[f1], p[u]); ///
        u = fa[f1], f1 = top[u];
    }
    if (u == v) return;
    if (deep[u] > deep[v]) swap(u, v); //使得deep[u] < deep[v]
    ne_update(1, p[son[u]], p[v]);
}

int query(int rt, int l, int r) { // 查询线段树中[l, r] 的最大值
    if (segTree[rt].l == l && segTree[rt].r == r)
        return segTree[rt].maxm;
    push_down(rt); ///
    int mid = (segTree[rt].l + segTree[rt].r) / 2;
    if (r <= mid) {
        return query(rt<<1, l, r);
    }else if (l > mid) {
        return query((rt<<1)|1, l, r);
    }else {
        return max(query(rt<<1, l, mid), query((rt<<1)|1, mid+1, r));
    }
    push_up(rt); ///
}


int findMax(int u, int v) {
    int f1 = top[u], f2 = top[v];
    int tmp = -100000000;
    while(f1 != f2) {
        if (deep[f1] < deep[f2]) {
            swap(f1, f2);
            swap(u, v);
        }
        tmp = max(tmp, query(1, p[f1], p[u]));
        u = fa[f1]; f1 = top[u];
    }
    if (u == v) return tmp;
    if (deep[u] > deep[v]) swap(u, v);
    return max(tmp, query(1, p[son[u]], p[v]));
}

int main() {
   // freopen("in.cpp", "r", stdin);
    int t;
    scanf("%d", &t);
    while(t--) {
        int n;
        scanf("%d", &n); // input
        init();
        for (int i=0; i<n; ++i) {
            scanf("%d%d%d", &e[i][0], &e[i][1], &e[i][2]);
            addEdge(e[i][0], e[i][1]);
            addEdge(e[i][1], e[i][0]);
        }
        dfs1(1, 0, 0);
        dfs2(1, 1);
        build(1, 0, pos-1);

        //线段树赋值
        for (int i=0; i<n-1; ++i) {
            if (deep[e[i][0]] > deep[e[i][1]]) {
                swap(e[i][0], e[i][1]);
            }
            update(1, p[e[i][1]], e[i][2]);
        }
        char op[10];
        int u, v;
        while(~scanf("%s", op)) {
//            printf("%s\n",op);
            if (op[0] == 'D') break;
            scanf("%d%d", &u, &v);
            if (op[0] == 'C') {
                update(1, p[e[u-1][1]], v);
            }else if (op[0] == 'N') {
                Negate(u, v);
            }else printf("%d\n", findMax(u, v));
        }
    }
    return 0;
}
View Code

 

HYSBZ 1036 树的统计Count

和上一题不同的是,这是点权,线段树维护的是每个点构成的数组。这个题因为只有单点修改,所以不需要延迟标记。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
using namespace std;

const int MAXN = 30010;

struct Edge
{
    int to,next;
}edge[MAXN*2];

int head[MAXN],tot;
int top[MAXN]; //top[v] 表示v所在的重链的顶端节点
int fa[MAXN]; //父亲节点
int deep[MAXN];//深度
int num[MAXN]; //num[v]表示以v为根的子树的节点数
int p[MAXN]; //p[v]表示v在线段树中的位置
int fp[MAXN];//和p数组相反
int son[MAXN];//重儿子
int pos;

void init()
{
    tot = 0;
    memset(head,-1,sizeof(head));
    pos = 0;
    memset(son,-1,sizeof(son));
}

void addedge(int u,int v)
{
    edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++;
}

void dfs1(int u,int pre,int d) //第一遍dfs求出fa,deep,num,son
{
    deep[u] = d;
    fa[u] = pre;
    num[u] = 1;
    for(int i = head[u];i != -1;i = edge[i].next)
    {
        int v = edge[i].to;
        if(v != pre)
        {
            dfs1(v,u,d+1);
            num[u] += num[v];
            if(son[u] == -1 || num[v] > num[son[u]])
                son[u] = v;
        }
    }
}

void getpos(int u,int sp)
{
    top[u] = sp;
    p[u] = pos++; ///
    fp[p[u]] = u; ///
    if(son[u] == -1) return;
    getpos(son[u],sp);
    for(int i = head[u]; i != -1 ; i = edge[i].next)
    {
        int v = edge[i].to;
        if(v != son[u] && v != fa[u]) getpos(v,v);
    }
}

struct Node
{
    int l,r;
    int sum;
    int Max;
}segTree[MAXN*3];

void push_up(int i)
{
    segTree[i].sum = segTree[i<<1].sum + segTree[(i<<1)|1].sum;
    segTree[i].Max = max(segTree[i<<1].Max,segTree[(i<<1)|1].Max);
}

int s[MAXN];

void build(int i,int l,int r)
{
    segTree[i].l = l;
    segTree[i].r = r;
    if(l == r)
    {
        segTree[i].sum = segTree[i].Max = s[fp[l]]; ///赋值
        return ;
    }
    int mid = (l + r)/2;
    build(i<<1,l,mid);
    build((i<<1)|1,mid+1,r);
    push_up(i);
}

void update(int i,int k,int val)//更新线段树的第k个值为val
{
    if(segTree[i].l == k && segTree[i].r == k)
    {
        segTree[i].sum = segTree[i].Max = val;
        return;
    }
    int mid = (segTree[i].l + segTree[i].r)/2;
    if(k <= mid)update(i<<1,k,val);
    else update((i<<1)|1,k,val);
    push_up(i);
}

int queryMax(int i,int l,int r)//查询线段树[l,r]区间的最大值
{
    if(segTree[i].l == l && segTree[i].r == r)
    {
        return segTree[i].Max;
    }
    int mid = (segTree[i].l + segTree[i].r)/2;
    if(r <= mid) return queryMax(i<<1,l,r);
    else if(l > mid)return queryMax((i<<1)|1,l,r);
    else return max(queryMax(i<<1,l,mid),queryMax((i<<1)|1,mid+1,r));
}

int querySum(int i,int l,int r) //查询线段树[l,r]区间的和
{
    if(segTree[i].l == l && segTree[i].r == r)
        return segTree[i].sum;
    int mid = (segTree[i].l + segTree[i].r)/2;
    if(r <= mid)return querySum(i<<1,l,r);
    else if(l > mid)return querySum((i<<1)|1,l,r);
    else return querySum(i<<1,l,mid) + querySum((i<<1)|1,mid+1,r);
}

int findMax(int u,int v)//查询u->v路径上节点的最大权值
{
    int f1 = top[u] , f2 = top[v];
    int tmp = -1000000000;
    while(f1 != f2)
    {
        if(deep[f1] < deep[f2])
        {
            swap(f1,f2);
            swap(u,v);
        }
        tmp = max(tmp,queryMax(1,p[f1],p[u]));
        u = fa[f1];
        f1 = top[u];
    }
    if(deep[u] > deep[v]) swap(u,v);
    return max(tmp,queryMax(1,p[u],p[v])); ///
}

int findSum(int u,int v) //查询u->v路径上节点的权值的和
{
    int f1 = top[u], f2 = top[v];
    int tmp = 0;
    while(f1 != f2)
    {
        if(deep[f1] < deep[f2])
        {
            swap(f1,f2);
            swap(u,v);
        }
        tmp += querySum(1,p[f1],p[u]);
        u = fa[f1];
        f1 = top[u];
    }
    if(deep[u] > deep[v]) swap(u,v);
    return tmp + querySum(1,p[u],p[v]); ///
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    int q;
    char op[20];
    int u,v;
    while(scanf("%d",&n) == 1)
    {
        init();
        for(int i = 1;i < n;i++)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        for(int i = 1;i <= n;i++)
            scanf("%d",&s[i]);
        dfs1(1,0,0);
        getpos(1,1);
        build(1,0,pos-1);
        scanf("%d",&q);
        while(q--)
        {
            scanf("%s%d%d",op,&u,&v);
            if(op[0] == 'C')
                update(1,p[u],v);//修改单点的值
            else if(strcmp(op,"QMAX") == 0)
                printf("%d\n",findMax(u,v));//查询u->v路径上点权的最大值
            else printf("%d\n",findSum(u,v));//查询路径上点权的和
        }
    }
    return 0;
}
View Code

参考链接:http://www.cnblogs.com/kuangbin/category/507663.html

posted on 2016-11-06 20:13  小小八  阅读(205)  评论(0编辑  收藏  举报