dfs序基本类型总结

大致看了下dfs序的题型，大致清楚了大致的解题思路。。。但是对于一些题目还是比较无力。。。。

dfs序比较重要的性质：一棵子树的所有节点在dfs序里是连续一段，主要就是利用这个性质来解题

题型一：对某个点X权值加上一个数W，查询某个子树X里所有点权值和。

解：列出dfs序，实现修改一个数，查询一段序列的和，显然这个序列可以用树状数组维护。

/*poj3321树状数组直接在第一次出现的位置+1，-1好了，对其他兄弟树没有影响，因为兄弟树是求区间，前面的+1，-1已经抵消掉了*/#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <string>#define mem(a,b) memset(a,b,sizeof(a))using namespace std;typedef long long ll;const int maxn = 100005;struct ppp{int v,nex;}e[maxn * 4] ;int head[maxn],tole,n;void make_edge(int u,int v){e[tole].v = v;e[tole].nex = head[u];head[u] = tole++;}int qian[maxn],hou[maxn],vis[maxn];int sum[maxn * 2];int cnt_node;void init(){for(int i = 0;i < 2 * maxn;i++){head[i >> 1] = -1;sum[i] = 0;vis[i >> 1] = 1;}tole = 0;cnt_node = 0;}void dfs(int u,int pre){qian[u] = ++cnt_node;for(int i = head[u];~i;i = e[i].nex){int v = e[i].v;if(v == pre)continue;dfs(v,u);}hou[u] = ++cnt_node;}int lowbit(int x){return x & -x;}int query(int x){int ret = 0;while(x >= 1){ret += sum[x];x -= lowbit(x);}return ret;}void add(int x,int v){while(x <= cnt_node){sum[x] += v;x += lowbit(x);}}int main(){while(~scanf("%d",&n)){init();for(int i = 1,a,b;i <= n - 1;i++){scanf("%d%d",&a,&b);make_edge(a,b);make_edge(b,a);}dfs(1,-1);for(int i = 1;i <= n;i++){add(qian[i],1);}int q;scanf("%d",&q);char c;int a;while(q--){scanf(" %c%d",&c,&a);if(c == 'Q'){int now = query(hou[a]) - query(qian[a] - 1);printf("%d\n",now);}else {if(vis[a] > 0){add(qian[a],-1);}else {add(qian[a],1);}vis[a] = !vis[a];}}}}

题型二：对X到Y的最短路上所有点权值加上一个数W，查询某个点的权值。

解：并没有找到该类型的题目。。。。假设有这种题目吧。。若X到Y上的点的权值都加上W，那么其实就是X到根的权值加上W，Y到根的点权值加上W，lca（X,Y）到根的权值减去W，parent(lca(X,Y))到根的点权值减去W。

#include<bits/stdc++.h>#define mem(a,b) memset(a,b,sizeof(a))#define FOR(i,a,b) for(int i = a;i <= b;i++)using namespace std;typedef long long ll;const int maxn = 100005;const int INF = 0x3f3f3f3f;struct ppp{int v,nex;}e[maxn * 2];int head[maxn],q,n;int dp[maxn * 2][19],tole;int dep[maxn],pos[maxn],last[maxn * 2];int cnt;int st[maxn],ed[maxn];void make_edge(int u,int v){e[tole].v = v,e[tole].nex = head[u];head[u] = tole++;}int sum[maxn * 2],parent[maxn];inline int lowbit (int x){return x & -x;}void add(int x,int v){while(x <= cnt){sum[x] += v;x += lowbit(x);}}int que(int x){int ret = 0;while(x >= 1){ret += sum[x];x -= lowbit(x);}return ret;}void dfs(int u,int fa,int deep){parent[u] = fa;dep[u] = deep;pos[u] = ++cnt;st[u] = cnt;last[cnt] = u;int v;for(int i = head[u];~i;i = e[i].nex){v = e[i].v;if(v == fa)continue;dfs(v,u,deep + 1);pos[u] = ++cnt;last[cnt] = u;}ed[u] = cnt;}void ST(){for(int i = 1;i <= cnt;i++)dp[i][0] = last[i];for(int j = 1;(1 << j) <= cnt;j++)for(int i = 1;i + (1 << j) - 1 <= cnt;i++){dp[i][j] = dp[i][j - 1];if(dep[dp[i + (1 << (j - 1))][j - 1]] < dep[dp[i][j - 1]])dp[i][j] = dp[i + (1 << (j - 1))][j - 1];}}int query(int l,int r){int k = 0;while((1 << (k + 1)) <= r - l + 1)k++;if(dep[dp[l][k]] < dep[dp[r - (1 << k) + 1][k]])return dp[l][k];else return dp[r - (1 << k) + 1][k];}void init(){mem(head,-1);tole = 0;cnt = 0;mem(sum,0);}int main(){while(~scanf("%d",&n)){init();for(int i = 1,a,b;i <= n - 1;i++){scanf("%d%d",&a,&b);make_edge(a,b);make_edge(b,a);}dfs(1,-1,1);ST();scanf("%d",&q);//查询个数char c;int a,b;while(q--){scanf(" %c",&c);if(c == 'Q'){scanf("%d",&a);printf("%d\n",que(ed[a]) - que(st[a] - 1));}else {int v;scanf("%d%d%d",&a,&b,&v);int aa = pos[a],bb = pos[b];if(aa > bb)swap(aa,bb);int lca = query(aa,bb);add(st[a],v);//这里在树状数组上加加减减感觉挺难理解的，可以写出dfs序画一画add(st[b],v);add(st[lca],-v);if(lca != 1)add(st[parent[lca]],-v);}}}}

题型三：

从X到Y的点加上或者减去一个W，求某个点子树内的所有点的权值和

假设X在Y的子树内，那么对于询问点Y，询问值会加上W * (depth(X) - depth(Y) + 1)。

拆开后为W * (depth(x) + 1) - W * (depth(Y)) 。因为对于询问的是Y，由此可见此式子跟X无关，所以可以开设两个树状数组，一个保存W * (depth(x) + 1)，另一个保存W * depth(Y)，对于每一个部分维护，维护的方式类似题型二。

以下代码非本人写，供参考

#include <bits/stdc++.h>#define mem(a,b) memset(a,b,sizeof(a))using namespace std;typedef long long ll;const int MAXN = 1e5+10;vector<int> edge[MAXN];int s[2*MAXN];int s1[2*MAXN];int seq[2*MAXN];int seq1[2*MAXN];int depth[2*MAXN];int first[MAXN];int dp[2*MAXN][25];int st[MAXN];int ed[MAXN];int parent[MAXN];int cnt, num;int Lowbit(int x){    return x & (-x);}void Add(int x, int val, int n){    if(x <= 0) return;    for(int i = x; i <= n; i += Lowbit(i)) {        s[i] += val;    }}void Add1(int x, int val, int n){    if(x <= 0) return;    for(int i = x; i <= n; i += Lowbit(i)) {        s1[i] += val;    }}int Sum(int x){    int res = 0;    for(int i = x; i > 0; i -= Lowbit(i)) {        res += s[i];    }    return res;}int Sum1(int x){    int res = 0;    for(int i = x; i > 0; i -= Lowbit(i)) {        res += s1[i];    }    return res;}void Dfs(int u, int fa, int dep){    parent[u] = fa;    seq[++cnt] = u;    seq1[++num] = u;    first[u] = num;    depth[num] = dep;    st[u] = cnt;    int len = edge[u].size();    for(int i = 0; i < len; i++) {        int v = edge[u][i];        if(v != fa) {            Dfs(v, u, dep+1);            seq1[++num] = u;            depth[num] = dep;        }    }    seq[++cnt] = u;    ed[u] = cnt;}void RMQ_Init(int n){    for(int i = 1; i <= n; i++) {        dp[i][0] = i;    }    for(int j = 1; (1 << j) <= n; j++) {        for(int i = 1; i + (1 << j) - 1 <= n; i++) {            int a = dp[i][j-1], b = dp[i + (1 << (j-1))][j-1];            dp[i][j] = depth[a] < depth[b] ? a : b;        }    }}int RMQ_Query(int l, int r){    int k = 0;    while((1 << (k + 1)) <= r - l + 1) k++;    int a = dp[l][k], b = dp[r-(1<<k)+1][k];    return depth[a] < depth[b] ? a : b;}int LCA(int u, int v){    int a = first[u], b = first[v];    if(a > b) a ^= b, b ^= a, a ^= b;    int res = RMQ_Query(a, b);    return seq1[res];}void Init(int n){    for(int i = 0; i <= n; i++) {        edge[i].clear();    }    memset(s, 0, sizeof(s));}int main(){    int n, op;    int u, v, w;    int cmd;    while(scanf("%d %d", &n, &op) != EOF) {        Init(n);        for(int i = 0; i < n-1; i++) {            scanf("%d %d", &u, &v);            edge[u].push_back(v);            edge[v].push_back(u);        }        cnt = 0, num = 0;        Dfs(1, -1, 0);        RMQ_Init(num);        while(op--) {            scanf("%d", &cmd);            if(cmd == 0) {                scanf("%d %d %d", &u, &v, &w);                int lca = LCA(u, v);//以下维护跟求值其实跟题型二是一样的，就是分两个树状数组                 Add(st[u], w * depth[first[u]] + w, cnt);                Add1(st[u], w, cnt);                Add(st[v], w * depth[first[v]] + w, cnt);                Add1(st[v], w, cnt);                Add(lca, -(w * depth[first[lca]] + w), cnt);                Add1(lca, -w, cnt);                Add(parent[lca], -(w * depth[first[parent[lca]]] + w), cnt);                Add1(parent[lca], -w, cnt);            }            else if(cmd == 1) {                scanf("%d", &u);                printf("%d\n", Sum(ed[u]) - Sum(st[u] - 1) - depth[first[u]] * (Sum1(ed[u]) - Sum1(st[u] - 1)));            }        }    }    return 0;}

题型四：对于某个节点X加上一个W，查询X到Y路径上的所有点的权值和。ps：这个感觉比前面的都容易想到啊。。。。

那么加上一个W的维护就跟题型一的维护差不多。

查询的值就等于：   X到根的权值和 + Y到跟的权值和 - LCA(X,Y)到根的权值和 - parent(LCA(X,Y))到根的权值和。

思路很简单，贴一下别人的代码：

const int MAXN = 1e5+10;vector<int> edge[MAXN];int s[2*MAXN];int s1[2*MAXN];int seq[2*MAXN];int seq1[2*MAXN];int depth[2*MAXN];int first[MAXN];int dp[2*MAXN][25];int st[MAXN];int ed[MAXN];int parent[MAXN];int cnt, num;int Lowbit(int x){    return x & (-x);}void Add(int x, int val, int n){    if(x <= 0) return;    for(int i = x; i <= n; i += Lowbit(i)) {        s[i] += val;    }}int Sum(int x){    int res = 0;    for(int i = x; i > 0; i -= Lowbit(i)) {        res += s[i];    }    return res;}void Dfs(int u, int fa, int dep){    parent[u] = fa;    seq[++cnt] = u;    seq1[++num] = u;    first[u] = num;    depth[num] = dep;    st[u] = cnt;    int len = edge[u].size();    for(int i = 0; i < len; i++) {        int v = edge[u][i];        if(v != fa) {            Dfs(v, u, dep+1);            seq1[++num] = u;            depth[num] = dep;        }    }    seq[++cnt] = u;    ed[u] = cnt;}void RMQ_Init(int n){    for(int i = 1; i <= n; i++) {        dp[i][0] = i;    }    for(int j = 1; (1 << j) <= n; j++) {        for(int i = 1; i + (1 << j) - 1 <= n; i++) {            int a = dp[i][j-1], b = dp[i + (1 << (j-1))][j-1];            dp[i][j] = depth[a] < depth[b] ? a : b;        }    }}int RMQ_Query(int l, int r){    int k = 0;    while((1 << (k + 1)) <= r - l + 1) k++;    int a = dp[l][k], b = dp[r-(1<<k)+1][k];    return depth[a] < depth[b] ? a : b;}int LCA(int u, int v){    int a = first[u], b = first[v];    if(a > b) a ^= b, b ^= a, a ^= b;    int res = RMQ_Query(a, b);    return seq1[res];}void Init(int n){    for(int i = 0; i <= n; i++) {        edge[i].clear();    }    memset(s, 0, sizeof(s));}int main(){    int n, op;    int u, v, w;    int cmd;    while(scanf("%d %d", &n, &op) != EOF) {        Init(n);        for(int i = 0; i < n-1; i++) {            scanf("%d %d", &u, &v);            edge[u].push_back(v);            edge[v].push_back(u);        }        cnt = 0, num = 0;        Dfs(1, -1, 0);        RMQ_Init(num);        while(op--) {            scanf("%d", &cmd);            if(cmd == 0) {                scanf("%d %d", &u, &w);                Add(st[u], w, cnt);                Add(ed[u], -w, cnt);            }            else if(cmd == 1) {                scanf("%d %d", &u, &v);                int lca = LCA(u, v);                printf("%d\n", Sum(st[u]) + Sum(st[v]) - Sum(st[lca]) - Sum(st[parent[lca]]));            }        }    }    return 0;}

题型五：对于一个点X的子树内的所有点加上一个值W，查询某个点的值。

那么只要在子树总都+w就好了，就是维护区间就是st[X] + w,ed[X] - w。

查询只要找que(X)就好了，注意如果点有原值的话要加上原值。

#include<bits/stdc++.h>#define mem(a,b) memset(a,b,sizeof(a))#define FOR(i,a,b) for(int i = a;i <= b;i++)using namespace std;typedef long long ll;const int maxn = 100005;const int INF = 0x3f3f3f3f;struct ppp{int v,nex;}e[maxn * 2];int head[maxn],q,n;int dp[maxn * 2][19],tole;int dep[maxn],pos[maxn],last[maxn * 2];int cnt;int st[maxn],ed[maxn];void make_edge(int u,int v){e[tole].v = v,e[tole].nex = head[u];head[u] = tole++;}int sum[maxn * 2],parent[maxn];inline int lowbit (int x){return x & -x;}void add(int x,int v){while(x <= cnt){sum[x] += v;x += lowbit(x);}}int que(int x){int ret = 0;while(x >= 1){ret += sum[x];x -= lowbit(x);}return ret;}void dfs(int u,int fa,int deep){parent[u] = fa;dep[u] = deep;pos[u] = ++cnt;st[u] = cnt;last[cnt] = u;int v;for(int i = head[u];~i;i = e[i].nex){v = e[i].v;if(v == fa)continue;dfs(v,u,deep + 1);pos[u] = ++cnt;last[cnt] = u;}ed[u] = cnt;}void ST(){for(int i = 1;i <= cnt;i++)dp[i][0] = last[i];for(int j = 1;(1 << j) <= cnt;j++)for(int i = 1;i + (1 << j) - 1 <= cnt;i++){dp[i][j] = dp[i][j - 1];if(dep[dp[i + (1 << (j - 1))][j - 1]] < dep[dp[i][j - 1]])dp[i][j] = dp[i + (1 << (j - 1))][j - 1];}}int query(int l,int r){int k = 0;while((1 << (k + 1)) <= r - l + 1)k++;if(dep[dp[l][k]] < dep[dp[r - (1 << k) + 1][k]])return dp[l][k];else return dp[r - (1 << k) + 1][k];}void init(){mem(head,-1);tole = 0;cnt = 0;mem(sum,0);}int main(){while(~scanf("%d",&n)){init();for(int i = 1,a,b;i <= n - 1;i++){scanf("%d%d",&a,&b);make_edge(a,b);make_edge(b,a);}dfs(1,-1,1);ST();scanf("%d",&q);//查询个数char c;int a,b;while(q--){scanf(" %c",&c);if(c == 'Q'){scanf("%d",&a);printf("%d\n",que(st[a]));//其中这个是更改后的值统计，如果有原值的话要加上原值}else {int w;scanf("%d%d",&a,&w);add(st[a],w);add(ed[a],-w);}}}}

题型六：对子树X里所有的节点加上一个值W，查询某个子树的权值和。

明显的区间修改，区间查询，那么。。。线段树。。

修改：就在st[X]到ed[X]上都加一个W

查询：直接查询st[X]到ed[X]，记住都到的值除以2，因为每个点都算了两遍

代码就不挂了。

题型七：对于子树X内的所有点都加上一个值W，查询X到Y之间的路径上所有点权和。

那么这题就跟题型四很像，就是修改的时候改成修改一段区间，从X到Y路径上的话直接就套用题型四的做法。

贴一下别人的代码：

typedef struct {    int l, r, sum, add;} Seg;const int MAXN = 1e5+10;Seg T[4*MAXN];vector<int> edge[MAXN];int s[2*MAXN];int s1[2*MAXN];int seq[2*MAXN];int seq1[2*MAXN];int depth[2*MAXN];int first[MAXN];int dp[2*MAXN][25];int parent[MAXN];int st[MAXN];int ed[MAXN];int cnt, cnt1;int Lowbit(int x){    return x & (-x);}void Add(int x, int val, int n){    if(x <= 0) return ;    for(int i = x; i <= n; i += Lowbit(i)) {        s[i] += val;    }}void Add1(int x, int val, int n){    if(x <= 0) return ;    for(int i = x; i <= n; i += Lowbit(i)) {        s1[i] += val;    }}int Sum(int x){    int res = 0;    for(int i = x; i > 0; i -= Lowbit(i)) {        res += s[i];    }    return res;}int Sum1(int x){    int res = 0;    for(int i = x; i > 0; i -= Lowbit(i)) {        res += s1[i];    }    return res;}void RMQ_Init(int n){    for(int i = 1; i <= n; i++) {        dp[i][0] = i;    }    for(int j = 1; (1 << j) <= n; j++) {        for(int i = 1; i + (1 << j) - 1 <= n; i++) {            int a = dp[i][j-1], b = dp[i + (1 << (j-1))][j-1];            dp[i][j] = depth[a] < depth[b] ? a : b;        }    }}int RMQ_Query(int l, int r){    int k = 0;    while((1 << (k + 1)) <= r - l + 1) k++;    int a = dp[l][k], b = dp[r-(1 << k)+1][k];    return depth[a] < depth[b] ? a : b;}int LCA(int u, int v){    int a = first[u], b = first[v];    if(a > b) a ^= b, b ^= a, a ^= b;    int res = RMQ_Query(a, b);    return seq1[res];}void Dfs(int u, int fa, int dep){    seq[++cnt] = u;    seq1[++cnt1] = u;    first[u] = cnt1;    parent[u] = fa;    depth[cnt1] = dep;    st[u] = cnt;    int len = edge[u].size();    for(int i = 0; i < len; i++) {        int v = edge[u][i];        if(v != fa) {            Dfs(v, u, dep+1);            seq1[++cnt1] = u;            depth[cnt1] = dep;        }    }    seq[++cnt] = u;    ed[u] = cnt;}void Init(int n){    for(int i = 0; i <= n; i++) {        edge[i].clear();    }    memset(s, 0, sizeof(s));    memset(s1, 0, sizeof(s1));}void Debug(){    int u, v;    while(1) {        scanf("%d %d", &u, &v);        printf("The LCA of %d %d is %d\n", u, v, LCA(u, v));    }}int main(){    int n, op;    int u, v, w;    int cmd;    while(scanf("%d %d", &n, &op) != EOF) {        Init(n);        for(int i = 0; i < n-1; i++) {            scanf("%d %d", &u, &v);            edge[u].push_back(v);            edge[v].push_back(u);        }        cnt = cnt1 = 0;        Dfs(1, 0, 0);        RMQ_Init(cnt1);        while(op--) {            scanf("%d", &cmd);            if(cmd == 0) {                scanf("%d %d", &u, &w);                Add(st[u], w * (1 - depth[first[u]]), cnt);                Add(ed[u], -w * (1 - depth[first[u]]), cnt);                Add1(st[u], w, cnt);                Add1(ed[u], -w, cnt);            }            else if(cmd == 1) {                scanf("%d %d", &u, &v);                int lca = LCA(u, v);                int par = parent[lca];                int ans = Sum(st[u]);                ans += depth[first[u]] * Sum1(st[u]);                ans += Sum(st[v]);                ans += depth[first[v]] * Sum1(st[v]);                ans -= Sum(st[lca]);                ans -= depth[first[lca]] * Sum1(st[lca]);                ans -= Sum(st[par]);                ans -= depth[first[par]] * Sum1(st[par]);                printf("%d\n", ans);            }        }    }    return 0;}