树的dfs序入门，BZOJ1103 ，hdu6162,EOJ3335

讨论昨天的02

克拉丽丝说不需要树剖可以直接dfs序

我不理解

他就丢给我这题，曰：经典的题目

百度题解一堆，好算知道dfs序是啥意思了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 5e5 + 7;
int l[N], r[N];

struct binaryIndexTree{
    int tree[N], n;
    inline void init(int n){
        this->n = n;
        memset(tree, 0, sizeof(tree));
    }
    inline void add(int k, int num){
        for (;k <= n; k += k&-k) tree[k] += num;
    }
    int sum(int k){
        int sum = 0;
        for (; k; k -= k&-k) sum += tree[k];
        return sum;
    }
} T;

struct graph{
    struct Edge{
        int from, to, nxt;
        Edge(){}
        Edge(int u, int v, int n):from(u), to(v), nxt(n){}
    }edges[N];
    int n, E, head[N];
    int top;

    inline void AddEdge(int f, int t){
        edges[++E] = Edge(f, t, head[f]);
        head[f] = E;
    }
    inline void Init(int n){
        this -> n = n ; E = -1; top = 0;
        for (int i = 0; i <= n; i++) head[i] = -1;
    }
    void dfs(int u){
        l[u] = ++top; T.add(top, 1);
        for (int i = head[u]; i != -1; i = edges[i].nxt){
            dfs(edges[i].to);
        }
        r[u] = ++top; T.add(top, -1);
    }
} g ;

int main(){
    //freopen("in.txt", "r", stdin);
    int n, m, u, v;
    char ch;
    for (; ~scanf("%d", &n);){
        g.Init(n);
        T.init(n*2);
        for (int i = 1; i < n; i++){
            scanf("%d%d", &u, &v);
            if (u > v) swap(u, v);
            g.AddEdge(u, v);
        }
        g.dfs(1);

        scanf("%d", &m);
        for (m += n-1; m--;){
            getchar();
            scanf("%c %d", &ch, &u);
            if (ch == 'W') printf("%d
", T.sum(l[u])-1);
            else {
                scanf("%d", &v);
                if (u > v) swap(u, v);
                T.add(l[v], -1); T.add(r[v], 1);
            }
        }
    }
    return 0;
}

写完之后，换了种实现昨天02的方法

dfs序配合之前写的离线查询

树状数组更加舒服

这里测试了一下效率

一开始在hdu似乎wa和MLE，后来发现fa[N][22]太大了，改成fa[N][18],

fa需要memset，一开始没有memset报WA，后来memset报MLE

改小了，memset才能Accept，也就是说，hdu的评测机，如果没有用到开到的内存，是不会爆MLE的

wa的原因是

        for (int i = LCADEP; i >= 0; i--) if (fa[x][i] != fa[y][i]){
            x = fa[x][i]; y = fa[y][i];
        }

如果前一组数据比后一组大，不memset fa的话，会因为前一组的傻逼值而算错lca

还有

本题原来北邮提供的数据是个巨型菊花图，章鱼哥嫌弃数据太弱了

加强了一波数据放在了EOJ3335上，各位非暴力选手可以试试了

非常艰难的在hdu上用树状数组过了之后，作死想用线段树，然后wa哭了

然而又一次轻松的在EOJ上过了，章鱼哥说EOJ是单组数据评测

ZKW线段树固然常数小，不过还是比树状数组差一点，很接近

下面这段代码，hdu会蜜汁wa，我已经不想管了，想在hdu上A掉，把线段树删了，树状数组保留就好，注释都有，模快化很容易改

#include <map>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1e5 + 7;
typedef long long LL;
int l[N], r[N];

struct gift{
    int pos;
    LL val;
    void read(int id){
        scanf("%lld", &val);
        pos = id;
    }
    bool operator < (const gift & b) const {
        return val < b.val;
    }
} gifts[N];

int ks[N * 2], K, H;
map<LL, int> hashK;
vector<int> whoAsk[N*2];
void insertK(int id, LL k){
    if (hashK.find(k) == hashK.end()) {
        hashK[k] = ++H;
        whoAsk[H].clear();
    }
    whoAsk[hashK[k]].push_back(id);
}
struct ask{
    int u, v, pos;
    LL a, b;
    vector<LL> ans;

    void read(int pos){
        this->pos = pos;
        ans.clear();
        scanf("%d%d%lld%lld", &u, &v, &a, &b);
        a--;
        ks[++K] = a, ks[++K] = b;
        insertK(pos, a);
        insertK(pos, b);
    }
    inline void print(){
        printf("%lld", abs(ans[1] - ans[0]));
    }
} asks[N];

struct binaryIndexTree{
    LL val[N * 2];
    int n;
    inline void build(int n){
        this->n = n;
        memset(val, 0, sizeof(val));
    }
    inline void add(int k, LL num){
        for (;k <= n; k += k&-k) val[k] += num;
    }
    LL sum(int k){
        if (k == 0) return 0;
        LL sum = 0;
        for (; k; k -= k&-k) sum += val[k];
        return sum;
    }
} TT ;

struct segTree{
    LL tree[N * 6];
    int M;
    inline void build(int n){
        M = 1; for(;M<n;) M<<=1; if(M!=1)M--;
        memset(tree, sizeof(tree), 0);
    }
    void add(int t, LL x){
        for (tree[t+=M]+=x, t>>=1; t; t>>=1){
            tree[t] = tree[t<<1] + tree[t<<1^1];
        }
    }
    LL sum(int l, int r){
        if (l > r || r == 0) return 0;
        LL ans = 0;
        for (l+=M-1,r+=M+1; l^r^1; l>>=1,r>>=1){
            if (~l&1) ans += tree[l^1];
            if ( r&1) ans += tree[r^1];
        }
        return ans;
    }
} T;

struct graph{
    struct Edge{
        int from, to, nxt;
        Edge(){}
        Edge(int u, int v, int n):from(u), to(v), nxt(n){}
    } edges[N * 2];
    static const int LCADEP = 17;
    int n, E, head[N];
    int top, dep[N], fa[N][LCADEP + 1];

    inline void AddEdge(int f, int t){
        edges[++E] = Edge(f, t, head[f]);
        head[f] = E;
    }
    inline void Init(int n){
        this -> n = n ; E = -1; top = 0; dep[0] = 0;
        for (int i = 0; i <= n; i++) head[i] = -1;
        memset(fa, 0, sizeof(fa));
    }

    void dfs(int u, int pre){
        l[u] = ++top;
        //printf("l[%d] = %d
", u, top);
        fa[u][0] = pre;
        dep[u] = dep[pre] + 1;
        for (int i = 1; i <= LCADEP; i++){
            if (dep[u] < (1<<i)) break;
            fa[u][i] = fa[fa[u][i-1]][i-1];
        }
        for (int i = head[u]; i != -1; i = edges[i].nxt){
            if (edges[i].to != pre) dfs(edges[i].to, u);
        }
        r[u] = ++top;
        //printf("r[%d] = %d
", u, top);
    }

    int lca(int x, int y){
        if (dep[x] < dep[y]) swap(x,y);
        int t = dep[x] - dep[y];
        for (int i = 0; i <= LCADEP; i++) if ((1<<i) & t) x = fa[x][i];
        for (int i = LCADEP; i >= 0; i--) if (fa[x][i] != fa[y][i]){
            x = fa[x][i]; y = fa[y][i];
        }
        return x==y ? x : fa[x][0];
    }

    void solve(ask &a){
        int u = a.u, v = a.v;
        int f = lca(u, v);
        LL ans = T.sum(1, l[u]) + T.sum(1, l[v]) - T.sum(1, l[f]) - T.sum(1, l[fa[f][0]]);
        //LL ans = T.sum(l[u]) + T.sum(l[v]) - T.sum(l[f]) - T.sum(l[fa[f][0]]);
        a.ans.push_back(ans);
    }
} g ;

int main () {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int n, m, u, v;
    for (; cin >> n >> m;) {
        for(int i = 1; i <= n; i++) gifts[i].read(i);
        sort(gifts + 1, gifts + n+1);
        g.Init(n);
        for(int i = 0; i < n - 1; i++) {
            scanf("%d%d", &u, &v);
            g.AddEdge(u, v);
            g.AddEdge(v, u);
        }
        g.dfs(1, 0);

        T.build(n*2);
        K = 0, H = 0;
        hashK.clear();
        for (int i = 1; i <= m; i++) asks[i].read(i);
        sort(ks + 1, ks + K+1);
        K = unique(ks + 1, ks + K+1) - (ks + 1);

        int cur = 1;
        for (int i = 1; i <= K; i++){
            //printf("ks[%d] = %d
", i, ks[i]);
            for (int &j = cur; j  <= n; j++){
                if (gifts[j].val > ks[i]) break;
                //printf("gifts[%d].val = %d, pos = %d, [%d, %d]
", j, gifts[j].val, gifts[j].pos, l[gifts[j].pos], r[gifts[j].pos]);
                T.add(l[gifts[j].pos], gifts[j].val);
                T.add(r[gifts[j].pos],-gifts[j].val);
            }
            int kk = hashK[ks[i]];
            for (int j = 0; j < whoAsk[kk].size(); j++){
                ask &a = asks[whoAsk[kk][j]];
                g.solve(a);
            }
        }

        for (int i = 1; i <= m; i++){
            asks[i].print();
            putchar(i==m ? '
' : ' ');
        }
    }
    return 0;
}