HDU 5957 线段树 附数据生成器

题目大意：

给一个树，然后再随便再多连一条边，形成一个带一个环的树。

一开始，所有树的节点的权重为0

有2个操作。

操作1： 距离结点u距离小于k的所有节点，权重+D

操作2：询问距离节点u距离小于k的所有节点的权重之和。

思路： 先假设题目的图只是一棵树，找一个根节点，BFS一下，dfn[i]表示i节点的BFS时间序列。 可以看出，每个节点的所有儿子的BFS时间序列是连续的，这就可以考虑到线段树来维护了。  并且，用线段树维护还挺好想的。

回到原题，这多一个环，把环弄出来（拓扑排序，最后度为2的点都是环上的）。

环上的点，视为若干个树的根。现在是好多棵树，每棵树的根彼此相连，依旧用线段树来解决，只不过在操作的时候，要多考虑一些东西~细节的话自己考虑一下就行了。

AC CODE:

#include <bits/stdc++.h>using namespace std;const int maxn =100000 + 100;vector<int>g[maxn];int n;void init(){scanf("%d", &n);for (int i = 0; i <= n; ++ i){g[i].clear();}for (int i = 1; i <= n; ++ i){int s, t;scanf("%d%d", &s, &t);g[s].push_back(t);g[t].push_back(s);}}const int maxnode = 6*maxn;struct node{int lson, rson;int val;//区间权重int change;//区间修改值,为0表示没有需要传递的}tree[maxnode];int tail;char input[20];int du[maxn];queue<int>q;int dfn[maxn], dfncnt;bool vis[maxn];vector<int>root[maxn];//作为根的节点的朋友节点void bfs(int arg){//du[i]==2表示为一个root//root[arg]可以知道arg作为根的时候，和哪些点直接连接root[arg].clear();for (auto will : g[arg])if (du[will] == 2)root[arg].push_back(will);//vis数组，是为了避免找儿子的时候，找到父亲了vis[arg] = true;q.push(arg);while (!q.empty()){int now = q.front();q.pop();dfn[now] = ++dfncnt;for (auto will : g[now]){if (du[will] == 2)continue;if (vis[will])continue;q.push(will);vis[will] = true;}}}int minson[maxn], maxson[maxn];//最小最大的儿子编号int mingrandson[maxn], maxgrandson[maxn];//最小最大的孙子编号int pre[maxn];//父节点//dfs是找出自己儿子层的数字，和孙子层的数字void dfs(int now, int fa){pre[now]=fa;minson[now] = mingrandson[now] = 0x7fffffff;maxson[now] = maxgrandson[now] = -1;for (auto will : g[now]){if (will == fa)continue;if (du[will] == 2)continue;dfs(will, now);if (dfn[will] > maxson[now])maxson[now] = dfn[will];if (dfn[will] < minson[now])minson[now] = dfn[will];if (minson[will] < mingrandson[now])mingrandson[now] = minson[will];if (maxson[will] > maxgrandson[now])maxgrandson[now] = maxson[will];}}//#define pr(x)cout<<#x<<" = "<<x<<" "//#define prln(x)cout<<#x<<" = "<<x<<endl#define pr(x)x#define prln(x)xvoid predoit(){memset(du,0,sizeof(du));memset(pre,-1,sizeof(pre));for (int i = 1; i <= n; ++ i){du[i] = g[i].size();if (du[i] == 1){//prln(i);q.push(i);}}while (!q.empty()){int now = q.front();q.pop();for (auto will : g[now]){du[will]--;if (du[will] == 1){q.push(will);}}}//for (int i = 1; i <= n; ++ i)//if (du[i] == 2)prln(i);memset(vis, 0, sizeof(vis));//访问标记memset(dfn, 0, sizeof(dfn));//BFS时间戳dfncnt = 0;//BFS时间戳记录器for (int i = 1; i <= n; ++ i)if (du[i] == 2){bfs(i);dfs(i, -1);}for (int i = 1; i <= n;++i)pr(i),prln(dfn[i]);}int ql, qr, qans;#define LSON tree[o].lson,L, M#define RSON tree[o].rson, M + 1 , R#define SELF o,L,R#define lc tree[o].lson#define rc tree[o].rsonvoid build(int o, int L, int R){tree[o].change = 0;tree[o].val = 0;if (L== R){tree[o].val = 0;return ;}tree[o].lson = ++ tail;tree[o].rson = ++ tail;int M = L + (R - L) / 2;build(LSON);build(RSON);}void push_down(int o, int L, int R)//向下传递标记{if (tree[o].change && L < R){tree[lc].change += tree[o].change;tree[rc].change += tree[o].change;tree[o].change = 0;}}void maintain(int o, int L, int R)//把o的儿子的信息，归到o节点{if (L < R){int M = L+(R-L)/2;tree[o].val = tree[lc].val + tree[rc].val + tree[lc].change * (M-L+1) + tree[rc].change * (R-M);}}int query(int o, int L, int R){if (ql <= L && R <= qr){return tree[o].val + tree[o].change * (R-L+1);//因为根节点的传递标记一直都在}push_down(SELF);int M = L + (R - L) / 2;int ret = 0;if (ql <= M)ret += query(LSON);else maintain(LSON);if (qr > M)ret += query(RSON);else maintain(RSON);maintain(SELF);return ret;}void update(int o, int L, int R){if (ql <= L && R <= qr){tree[o].change += qans;return;}else{push_down(SELF);int M = L + (R-L)/2;if (ql <= M)update(LSON);else maintain(LSON);if (qr > M)update(RSON);else maintain(RSON);}maintain(SELF);}vector<int>sb;void UD(int u, int k, int d){//一会儿赋值下QU，然后略改一下参数即可qans=d;if (!k){ql = dfn[u], qr = dfn[u];update(0, 1, n);pr(ql),prln(qr);}else if (k==1){ql = minson[u],qr = maxson[u];update(0, 1, n);pr(ql),prln(qr);if (du[u]!=2)//u非根{ql = qr = dfn[pre[u]];update(0, 1, n);pr(ql),prln(qr);ql = qr = dfn[u];update(0, 1, n);pr(ql),prln(qr);}else{//u为根for (auto will : root[u]){ql = qr = dfn[will];update(0, 1, n);pr(ql),prln(qr);}ql = qr = dfn[u];update(0, 1, n);pr(ql),prln(qr);}}else//也就是k==2{ql = minson[u], qr = maxson[u];update(0, 1, n);pr(ql),prln(qr);ql = mingrandson[u], qr = maxgrandson[u];update(0, 1, n);pr(ql),prln(qr);if (du[u]!=2){int fa = pre[u];ql = minson[fa], qr = maxson[fa];update(0, 1, n);pr(ql),prln(qr);ql = dfn[fa], qr = dfn[fa];update(0, 1, n);pr(ql),prln(qr);if (du[fa] != 2)//依旧没到环上{fa = pre[fa];ql = dfn[fa], qr = dfn[fa];update(0, 1, n);pr(ql),prln(qr);}else{for (auto will : root[fa]){ql = qr = dfn[will];update(0, 1, n);pr(ql),prln(qr);}}}else{sb.clear();for (auto will : root[u]){sb.push_back(will);ql = minson[will], qr = maxson[will];update(0, 1, n);pr(ql),prln(qr);for (auto x : root[will])sb.push_back(x);}sort(sb.begin(), sb.end());sb.erase(unique(sb.begin(),sb.end()),sb.end());for (auto will : sb){ql = qr = dfn[will];update(0, 1, n);pr(ql),prln(qr);}}}}void QU(int u, int k){//一会儿赋值下QU，然后略改一下参数即可int sum = 0;if (!k){ql = dfn[u], qr = dfn[u];sum += query(0, 1, n);pr(ql),prln(qr);}else if (k==1){ql = minson[u],qr = maxson[u];sum += query(0, 1, n);pr(ql),prln(qr);if (du[u]!=2)//u非根{ql = qr = dfn[pre[u]];sum += query(0, 1, n);pr(ql),prln(qr);ql= qr = dfn[u];sum += query(0, 1, n);pr(ql),prln(qr);}else{//u为根for (auto will : root[u]){ql = qr = dfn[will];sum += query(0, 1, n);pr(ql),prln(qr);}ql = qr = dfn[u];sum += query(0, 1, n);pr(ql),prln(qr);}}else//也就是k==2{ql = minson[u], qr = maxson[u];sum += query(0, 1, n);pr(ql),prln(qr);ql = mingrandson[u], qr = maxgrandson[u];sum += query(0, 1, n);pr(ql),prln(qr);if (du[u]!=2){int fa = pre[u];ql = minson[fa], qr = maxson[fa];sum += query(0, 1, n);pr(ql),prln(qr);ql = dfn[fa], qr = dfn[fa];sum += query(0, 1, n);pr(ql),prln(qr);if (du[fa] != 2)//依旧没到环上{fa = pre[fa];ql = dfn[fa], qr = dfn[fa];sum += query(0, 1, n);pr(ql),prln(qr);}else{for (auto will : root[fa]){ql = qr = dfn[will];sum += query(0, 1, n);pr(ql),prln(qr);}}}else{sb.clear();for (auto will : root[u]){sb.push_back(will);ql = minson[will], qr = maxson[will];sum += query(0, 1, n);pr(ql),prln(qr);for (auto x : root[will])sb.push_back(x);}sort(sb.begin(), sb.end());sb.erase(unique(sb.begin(),sb.end()),sb.end());for (auto will : sb){ql = qr = dfn[will];sum += query(0, 1, n);pr(ql),prln(qr);}}}printf("%d\n", sum);}void doit(){int T;int u, k, d;tail=0;//root为0build(0,1,n);//for (int o = 0; o <= tail; ++ o)//pr(lc),pr(rc),pr(tree[o].val),prln(tree[o].change);scanf("%d", &T);while (T--){scanf("%s", input);if (input[0]=='M'){scanf("%d%d%d", &u, &k, &d);UD(u,k,d);for (int o = 0; o <= tail; ++ o)pr(o),pr(lc),pr(rc),pr(tree[o].val),prln(tree[o].change);prln("\n\n"),prln("\n\n");}else{scanf("%d%d", &u, &k);QU(u, k);for (int o = 0; o <= tail; ++ o)pr(o),pr(lc),pr(rc),pr(tree[o].val),prln(tree[o].change);prln("\n\n"),prln("\n\n");}}}int main(){int T;scanf("%d", &T);while (T--){init();predoit();for (int i = 1; i <= n; ++ i)pr(minson[i]),pr(maxson[i]),pr(mingrandson[i]),prln(maxgrandson[i]);doit();//break;//暂且只测一组数据}return 0;}

data maker:

#include <cstdio>#include <cstring>#include <vector>#include <ctime>#include <iostream>#include <cstdlib>using namespace std;int g[500][500];int main(){srand(time(0));int T = 5;cout<<T<<endl;while (T--){memset(g,0,sizeof(g));int n = rand() % 10 + 3;cout<<n<<endl;for (int i = 2; i <= n; ++ i){int a = i, b = rand()%(i-1)+1;g[a][b]=g[b][a]=1;cout << a <<" " << b<<endl;}while (1){int a = rand()%n+1;int b = rand()%n+1;if (a!=b && !g[a][b]){cout<<a<<" "<<b<<endl;break;}}int ask=rand()%3+1;cout<<ask<<endl;while (ask --){string a[2];a[0]="MODIFY";a[1]="QUERY";int flag = rand()%2;if (flag){cout<<a[0]<<" "<<rand()%n+1<<" "<<rand()%2+1<<" "<<rand()%5<<endl;}else{cout<<a[1]<<" "<<rand()%n+1<<" "<<rand()%2+1<<endl;}}}return 0;}/*262 13 24 25 16 24 52QUERY 6 2MODIFY 6 2 372 13 14 15 26 37 26 12MODIFY 2 1 4QUERY 2 2*/