bupt summer training for 16 #5 ——数据结构

https://vjudge.net/contest/173780

A.假设 Pt = i，则由Ppi = i得

Ppt = t = Pi

所以就有 if Pt = i then Pi = t

 #include <cstdio>
 #include <algorithm>

 using namespace std;

 int n, a[];

 int main() {
     scanf("%d", &n);
     if(n & ) puts("-1");
     else {
         for(int i = ;i <= n;i ++)
             a[i] = i;
         for(int i = ;i <= n;i += )
             swap(a[i], a[i + ]);
         for(int i = ;i <= n;i ++)
             printf("%d ", a[i]);
     }
 }

B.

C.treap练手题，STL_set练手题，链表练手题

set用的不熟所以直接上的链表

排序后时间倒序来看的思路

 #include <cstdio>
 #include <algorithm>

 using namespace std;

 typedef long long ll;

 const int maxn = ;

 int n, b[maxn];

 ll ans;

 struct List{
     int next, last;
 }c[maxn];

 struct node{
     ll x;
     int y;
     bool operator < (const node &a) const {
         return x < a.x;
     }
 }a[maxn];

 int main() {
     scanf("%d", &n);
     for(int i = ;i <= n;i ++)
         scanf("%lld", &a[i].x), a[i].y = i;
     ans = a[].x;
     a[].x = -1000000000000ll;
     a[n + ].x = 1000000000000ll;
     sort(a + , a + n + );
     for(int i = ;i <= n;i ++) b[a[i].y] = i;
     for(int i = ;i <= n;i ++) {
         c[i].next = i + ;
         c[i].last = i - ;
     }
     for(int i = n;i > ;i --) {
         ans += min(a[b[i]].x - a[c[b[i]].last].x, a[c[b[i]].next].x - a[b[i]].x);
         c[c[b[i]].last].next = c[b[i]].next;
         c[c[b[i]].next].last = c[b[i]].last;
     }
     printf("%lld\n", ans);
     return ;
 }

D.边权给点，点修改和路径查询

直接上树剖+线段树，考场写狗

多组数据需要clear存边的vector，fa 和 son

修改的时候bel[i]找到第 i 条边权给的点

还要对应该点在线段树上的叶节点位置！

 #include <cstdio>
 #include <vector>
 #include <algorithm>

 using namespace std;

 const int maxn = ;

 int Case, n, M;

 struct Edge{
     int v, w, num;
 };

 vector <Edge> edge[maxn];

 char str[];

 int a, b, tr[maxn << ];

 int bel[maxn], siz[maxn], son[maxn], val[maxn];

 int cnt, fa[maxn], dep[maxn], pos[maxn], dfn[maxn], top[maxn];

 void dfs1(int u) {
     siz[u] = ;
     for(int v, i = ;i < edge[u].size();i ++) {
         v = edge[u][i].v;
         if(v == fa[u]) continue;
         fa[v] = u;
         dep[v] = dep[u] + ;
         val[v] = edge[u][i].w;
         bel[edge[u][i].num] = v;
         dfs1(v), siz[u] += siz[v];
         if(siz[son[u]] < siz[v]) son[u] = v;
     }
 }

 void dfs2(int u, int tp) {
     top[u] = tp, pos[u] = ++ cnt, dfn[cnt] = u;
     if(son[u]) dfs2(son[u], tp);
     for(int v, i = ;i < edge[u].size();i ++)  {
         v = edge[u][i].v;
         if(v != fa[u] && v != son[u])
             dfs2(v, v);
     }
 }

 void change(int i, int x) {
     i = pos[i];
     for(tr[i += M] = x, i >>= ;i;i >>= )
         tr[i] = max(tr[i << ], tr[i <<  | ]);
 }

 int mmax(int s, int t) {
     int res = -;
     for(s += M - , t += M + ;s ^ t ^ ;s >>= , t >>= ) {
         if(~ s & ) res = max(res, tr[s ^ ]);
         if(  t & ) res = max(res, tr[t ^ ]);
     }
     return res;
 }

 void query(int u, int v) {
     int fu = top[u], fv = top[v], res = -;
     while(fu != fv) {
         if(dep[fu] < dep[fv]) swap(u, v), swap(fu, fv);
         res = max(res, mmax(pos[fu], pos[u]));
         u = fa[fu], fu = top[u];
      }
      if(u == v) printf("%d\n", res);
      else {
          if(dep[u] < dep[v]) swap(u, v);
          printf("%d\n", max(res, mmax(pos[v] + , pos[u])));
      }
 }

 int main() {
 //freopen("test.txt","r",stdin);
     val[] = -;
     int u, v, w;
     scanf("%d", &Case);
     while(Case --) {
         cnt = ;
         scanf("%d", &n);
         for(int i = ;i <= n;i ++) edge[i].clear(), fa[i] = son[i] = ;
         for(int i = ;i < n;i ++) {
             scanf("%d %d %d", &u, &v, &w);
             edge[u].push_back((Edge){v, w, i});
             edge[v].push_back((Edge){u, w, i});
         }
         dfs1(), dfs2(, );
         for(M = ;M < n + ; M <<= );
         for(int i = ;i <= n;i ++) tr[i + M] = val[dfn[i]];
         for(int i = n + ;i <= M + ;i ++) tr[i + M] = val[];
         for(int i = M;i;i --) tr[i] = max(tr[i << ], tr[i <<  | ]);
         while(scanf("%s", str), str[] != 'D') {
             if(str[] == 'C') scanf("%d %d", &a, &b), change(bel[a], b);
             else scanf("%d %d", &a, &b), query(a, b);
         }
         puts("");
     }
 }

E.考场上脑补出来一个正解，可惜只差了一步

两个单调队列，一增一减，从前向后扫

当扫到某个位置，最大差距 > R 时

踢出两个队列中位置最靠前的元素直至差距 <= R

如果这是最大差距 >= L 则更新答案即可

时间复杂度O(n)，别人说这是个做过的RMQ啊...

 #include <cstdio>
 #include <algorithm>

 using namespace std;

 const int maxn = ;

 int n, x, y, m, a[maxn], q1[maxn], q2[maxn];

 int l1, r1, l2, r2, t;

 int main() {
     while(scanf("%d %d %d", &n, &x, &y) != EOF) {
         for(int i = ;i <= n;i ++)
             scanf("%d", &a[i]);
         m = t = , l1 = l2 = , r1 = r2 = ;
         for(int i = ;i <= n;i ++) {
             while(l1 <= r1 && a[i] > a[q1[r1]]) r1 --;
             q1[++ r1] = i;
             while(l2 <= r2 && a[i] < a[q2[r2]]) r2 --;
             q2[++ r2] = i;
             while(l1 <= r1 && l2 <= r2 && a[q1[l1]] > a[q2[l2]] + y) {
                 if(q1[l1] < q2[l2]) t = q1[l1 ++];
                 else t = q2[l2 ++];
             }
             if(l1 <= r1 && l2 <= r2 && a[q1[l1]] >= a[q2[l2]] + x) m = max(m, i - t);
         }
         printf("%d\n", m);
     }
     return ;
 }

F.就是枚举min，然后单调队列求出左右拓展多远就好了

 #include <cstdio>

 const int maxn = ;

 int l = , r, q[maxn];

 int L, R, n, a[maxn], b[maxn], c[maxn];

 long long ans = -, tmp, s[maxn];

 int main() {
     scanf("%d", &n);
     for(int i = ;i <= n;i ++)
         scanf("%d", &a[i]), s[i] = a[i] + s[i - ];
     for(int i = ;i <= n;i ++) {
         while(l <= r && a[i] < a[q[r]]) b[q[r]] = i - q[r], r --;
         q[++ r] = i;
     }
     while(l <= r) b[q[r]] = n +  - q[r], r --;
     l = , r = ;
     for(int i = n;i;i --) {
         while(l <= r && a[i] < a[q[r]]) c[q[r]] = q[r] - i, r --;
         q[++ r] = i;
     }
     while(l <= r) c[q[r]] = q[r], r --;
     for(int i = ;i <= n;i ++) {
         tmp = (s[i + b[i] - ] - s[i - c[i]]) * a[i];
         if(tmp > ans) ans = tmp, L = i - c[i] + , R = i + b[i] - ;
     }
     printf("%lld\n%d %d", ans, L ,R);
     return ;
 }

G.开始没看清题发现大家都会写区间众数？

连续的相同...线段树傻逼题

 #include <cstdio>
 #include <algorithm>

 using namespace std;

 const int maxn = ;

 struct node{
     int ls, rs, lc, rc, mm;
 }tr[maxn << ];

 int n, m, c[maxn];

 node operator +(const node &a, const node &b) {
     node res;
     res.ls = a.ls;
     res.lc = a.lc;
     res.rs = b.rs;
     res.rc = b.rc;
     res.mm = max(a.mm, b.mm);
     if(a.rc == b.lc) {
         res.mm = max(res.mm, a.rs + b.ls);
         if(a.lc == a.rc) res.ls += b.ls;
         if(b.lc == b.rc) res.rs += a.rs;
     }
     return res;
 }

 void build(int o, int l, int r) {
     if(l == r) {
         tr[o] = (node){, , c[r], c[r], };
         return;
     }
     int mid = (l + r) >> ;
     build(o << , l, mid);
     build(o <<  | , mid + , r);
     tr[o] = tr[o << ] + tr[o <<  | ];
 }

 node ask(int o, int l, int r, int s, int t){
     if(s <= l && r <= t) return tr[o];
     int mid = (l + r) >> ;
     if(t <= mid) return ask(o << , l, mid, s, t);
     else if(s > mid) return ask(o <<  | , mid + , r, s, t);
     else return ask(o << , l, mid, s, t) + ask(o <<  | , mid + , r, s, t);
 }

 int main() {
     int a, b;
     while(scanf("%d %d", &n, &m), n != ) {
         for(int i = ;i <= n;i ++)
             scanf("%d", &c[i]);
         build(, , n);
         //printf("%d %d %d\n",tr[1].mm,tr[6].rc, tr[7].lc);
         while(m --) {
             scanf("%d %d", &a, &b);
             printf("%d\n", ask(, , n, a, b).mm);
         }
     }
 }

H.

I.无修改查询树上路径和，随便做啦

dis[u] + dis[v] - dis[lca] * 2

我直接上的bfs树剖求lca

 #include <cstdio>
 #include <vector>
 #include <algorithm>

 using namespace std;

 const int maxn = ;

 struct Edge{
     int v, w;
 };

 vector <Edge> edge[maxn];

 int Case, n, m, a, b, c, dis[maxn];

 int l, r, q[maxn], siz[maxn], son[maxn];

 int cnt, fa[maxn], dep[maxn], top[maxn];

 int lca(int u, int v) {
     int fu = top[u], fv = top[v];
     while(fu != fv) {
         if(dep[fu] < dep[fv]) swap(u, v), swap(fu, fv);
         u = fa[fu], fu = top[u];
      }
      return dep[u] < dep[v] ? u : v;
 }

 int main() {
     int u, v, w;
     scanf("%d", &Case);
     while(Case --) {
         scanf("%d %d", &n, &m);
         for(int i = ;i < n;i ++) {
             scanf("%d %d %d", &u, &v, &w);
             edge[u].push_back((Edge){v, w});
             edge[v].push_back((Edge){u, w});
         }
         l = r = , q[++ r] = ;
         while(l <= r) {
             u = q[l ++];
             siz[u] = , son[u] = top[u] = ;
             for(int i = ;i < edge[u].size();i ++) {
                 v = edge[u][i].v;
                 if(v == fa[u]) continue;
                 fa[v] = u, q[++ r] = v, dep[v] = dep[u] + , dis[v] = dis[u] + edge[u][i].w;
             }
         }
         for(int i = n;i > ;i --) {
             siz[fa[q[i]]] += siz[q[i]];
             if(!son[fa[q[i]]] || siz[son[fa[q[i]]]] < siz[q[i]]) son[fa[q[i]]] = q[i];
         }
         for(int i = ;i <= n;i ++) {
             u = q[i];
             if(son[fa[u]] == u) top[u] = top[fa[u]];
             else top[u] = u;
         }
         while(m --) {
             scanf("%d %d", &a, &b);
             c = lca(a, b);
             printf("%d\n", dis[a] - dis[c] - dis[c] + dis[b]);
         }
     }
 }

J.裸树状数组

 #include <cstdio>

 const int maxn = ;

 int c[maxn], a[maxn];

 int Case, n, x, y;

 char str[];

 int lowbit(int x) {
     return (x & (-x));
 }

 int ask(int i) {
     int res = ;
     while(i > ) res += c[i], i -= lowbit(i);
     return res;
 }

 void add(int i, int x) {
     while(i <= n) c[i] += x, i += lowbit(i);
 }

 int main() {
     //freopen("test.txt", "r", stdin);
     scanf("%d", &Case);
     for(int tt = ;tt <= Case;tt ++) {
         printf("Case %d:\n", tt);
         scanf("%d", &n);
         for(int i = ;i <= n;i ++)
             scanf("%d", &a[i]), a[i] += a[i - ];
         while(scanf("%s", str), str[] != 'E') {
             scanf("%d %d", &x, &y);
             if(str[] == 'Q') printf("%d\n", ask(y) - ask(x - ) + a[y] - a[x - ]);
             else add(x, y * (str[] == 'A' ?  : -));
         }
         for(int i = ;i <= n;i ++) c[i] = ;
     }
     return ;
 }

K.区间加和区间求和，差分树状数组除非有封装好的板子

不然现场还是只能写线段树

 #include <cstdio>
 #include <algorithm>

 using namespace std;

 typedef long long ll;

 const int maxn = ;

 int n, m, d[maxn];

 char str[];

 int a, b, c;

 ll s[maxn];

 struct node{
     ll lazy, sum;
 }tr[maxn << ];

 void pushup(int o) {
     tr[o].sum = tr[o << ].sum + tr[o <<  | ].sum;
 }

 void pushdown(int o, int l, int r) {
     if(!tr[o].lazy) return;
     int mid = (l + r) >> ;
     tr[o << ].sum += tr[o].lazy * (mid - l + );
     tr[o <<  | ].sum += tr[o].lazy * (r - mid);
     tr[o << ].lazy += tr[o].lazy;
     tr[o <<  |].lazy += tr[o].lazy;
     tr[o].lazy = ;
 }

 void add(int o, int l, int r, int s, int t, int x) {
     if(l != r) pushdown(o, l, r);
     if(s <= l && r <= t) {
         tr[o].lazy += x;
         tr[o].sum += 1ll * x * (r - l + );
         return;
     }
     int mid = (l + r) >> ;
     if(s <= mid) add(o << , l, mid, s, t, x);
     if(mid < t) add(o <<  | , mid + , r, s, t, x);
     pushup(o);
 }

 ll query(int o, int l, int r, int s, int t) {
     if(l != r) pushdown(o, l, r);
     if(s <= l && r <= t) return tr[o].sum;
     ll res = ;
     int mid = (l + r) >> ;
     if(s <= mid) res += query(o << , l, mid, s, t);
     if(mid < t) res += query(o <<  | , mid + , r, s, t);
     pushup(o);
     return res;
 }

 int main(int argc, char const *argv[]){
     scanf("%d %d", &n, &m);
     for(int i = ;i <= n;i ++)
         scanf("%d", &d[i]), s[i] = s[i - ] + d[i];
     while(m --) {
         scanf("%s %d %d", str, &a, &b);
         if(str[] == 'Q') printf("%lld\n", query(, , n, a, b) + s[b] - s[a - ]);
         else scanf("%d", &c), add(, , n, a, b, c);
     }
     return ;
 }

L.vector会TLE改邻接表能过...***

dfs序 + 树状数组

 #include <cstdio>

 const int maxn = ;

 int n, m, q, a[maxn], c[maxn], st[maxn], en[maxn];

 int tot, head[maxn], to[maxn], next[maxn];

 void ade(int x, int y) {
     to[++ tot] = y, next[tot] = head[x], head[x] = tot;
 }

 int lowbit(int x) {
     return (x & (-x));
 }

 int ask(int i) {
     int res = ;
     while(i > ) res += c[i], i -= lowbit(i);
     return res;
 }

 void add(int i, int x) {
     while(i <= n) c[i] += x, i += lowbit(i);
 }

 void dfs(int x, int f) {
     a[x] = , st[x] = ++ m;
     for(int y, i = head[x];i;i = next[i]) {
         y = to[i];
         if(y == f) continue;
         dfs(y, x);
     }
     en[x] = m;
 }

 char str[];

 int x;

 int main() {
     scanf("%d", &n);
     for(int u, v, i = ;i < n;i ++) {
         scanf("%d %d", &u, &v);
         ade(u, v), ade(v, u);
     }
     dfs(, );
     scanf("%d", &q);
     while(q --) {
         scanf("%s %d", str, &x);
         if(str[] == 'Q') printf("%d\n", ask(en[x]) - ask(st[x] - ) + en[x] - st[x] + );
         else {
             if(a[x]) add(st[x], -), a[x] = ;
             else add(st[x], ), a[x] = ;
         }
     }
     return ;
 }