NOIP模拟31
作者：互联网
T1：
将问题分层考虑，显然我们可以求出最大得分，利用
前趋贪心即可，于是考虑在最大得分下，如何使字典序最
大，显然我们可以考虑在换牌的情况下对最大得分有无影
响，发现其具有单调性，于是在使用线段树维护最大得分
进行check的同时二分满足条件的最大得分即可
1 #include <bits/stdc++.h>
2 using namespace std;
3 #define I int
4 #define V void
5 const I N = 1e5 + 5;
6 I n;
7 struct SGT { I a[N << 2],b[N << 2],s[N << 2];
8 #define lid id << 1
9 #define rid id << 1 | 1
10 #define mid (l + r >> 1)
11 inline V update (I id) {
12 I pub (min (a[rid],b[lid]));
13 a[id] = a[lid] + a[rid] - pub;
14 b[id] = b[lid] + b[rid] - pub;
15 s[id] = s[lid] + s[rid] + pub;
16 }
17 V poimod_A (I id,I l,I r,I pos,I key) {
18 if (l == r) return (V) (a[id] += key);
19 pos <= mid ? poimod_A (lid,l,mid,pos,key) : poimod_A (rid,mid + 1,r,pos,key);
20 update (id);
21 }
22 V poimod_B (I id,I l,I r,I pos,I key) {
23 if (l == r) return (V) (b[id] += key);
24 pos <= mid ? poimod_B (lid,l,mid,pos,key) : poimod_B (rid,mid + 1,r,pos,key);
25 update (id);
26 }
27 #undef mid
28 #undef lid
29 #undef rid
30 }SGT;
31 signed main () { I a[N],b[N]; multiset <I> s;
32 cin >> n;
33 for (I i(1);i <= n; ++ i) cin >> b[i], SGT.poimod_B (1,1,N,b[i],1);
34 for (I i(1);i <= n; ++ i) cin >> a[i], SGT.poimod_A (1,1,N,a[i],1), s.insert (a[i]);
35 I tmp (SGT.s[1]);
36 for (I i(1);i <= n; ++ i) {
37 SGT.poimod_B (1,1,N,b[i],-1);
38 I l(b[i] + 1), r(*--s.end ());
39 while (l < r) { I mid (l + r + 1 >> 1);
40 SGT.poimod_A (1,1,N,mid,-1);
41 SGT.s[1] + 1 == tmp ? l = mid : r = mid - 1;
42 SGT.poimod_A (1,1,N,mid,1);
43 }
44 SGT.poimod_A (1,1,N,l,-1);
45 if (l <= r && SGT.s[1] + 1 == tmp) { printf ("%d ",l); -- tmp; s.erase (s.find (l)); }
46 else {
47 SGT.poimod_A (1,1,N,l,1);
48 I l(1), r(b[i]);
49 while (l < r) { I mid (l + r + 1 >> 1);
50 SGT.poimod_A (1,1,N,mid,-1);
51 SGT.s[1] == tmp ? l = mid : r = mid - 1;
52 SGT.poimod_A (1,1,N,mid,1);
53 }
54 SGT.poimod_A (1,1,N,l,-1);
55 printf ("%d ",l);
56 s.erase (s.find (l));
57 }
58 }
59 }
View Code
分层考虑，线段树维护辅助二分
T2：
发现问题具有明显的子结构，即无论最大值在哪，最小
值都必须移动到左右两侧，再次之后，次小值将继续这一进
程，故发现只要不断重复这一进程即可。
维护权值与距离边界距离，采取最优决策即可
1 #include <bits/stdc++.h>
2 using namespace std;
3 #define I int
4 #define V void
5 const I N = 1e5 + 5;
6 I n;
7 struct BIT { I c[N];
8 #define lowbit(x) (x & -x)
9 inline V insert (I x,I y) { for ( ;x <= n;x += lowbit(x)) c[x] += y; }
10 inline I secsum (I x) { I res(0);
11 for ( ; x ;x -= lowbit(x)) res += c[x];
12 return res;
13 }
14 }BIT;
15 signed main() { I upd,res; deque <I> q[N];
16 cin >> n;
17 for (I i(1),data;i <= n; ++ i) cin >> data, q[data].push_back(i), BIT.insert (i,1), upd = max (upd,data);
18 for (I i(1);i <= upd; ++ i) while (q[i].size ()) {
19 I x(q[i].front ()), y(q[i].back ()), tmp1(BIT.secsum (x - 1)), tmp2(BIT.secsum (n) - BIT.secsum (y));
20 tmp1 < tmp2 ? (res += tmp1, BIT.insert (x,-1), q[i].pop_front ()) : (res += tmp2, BIT.insert (y,-1), q[i].pop_back ());
21 }
22 cout << res << endl;
23 }
View Code
问题子结构性质，维护与最优决策
T3：
包含或不相交，显然在提示树形结构，于是可以进行树
形DP，发现并不必要，可以转化到序列问题进行考虑，DP
很显然，然而复杂度过不去，考虑优化。
发现问题中要求最大美观程度，于是可以转化DP，采取
差分形势进行优化，因为在最优条件下，差分数组必然单调
不降，于是DP过程中可以利用大根堆进行整体转移，相较于
传统背包DP，其省略了背包容量的枚举，因为在所有转移满
足最大性的情况下可以直接合并转移
1 #include <bits/stdc++.h>
2 using namespace std;
3 #define I int
4 #define V void
5 #define LL long long
6 const I N = 3e5 + 5;
7 I n,m,val[N],tot,head[N],to[N],nxt[N];
8 struct LINE {
9 I id,l,r;
10 friend inline bool operator < (const LINE &a,const LINE &b) {
11 return a.l == b.l ? a.r > b.r : a.l < b.l;
12 }
13 }line[N];
14 LL sup[N];
15 priority_queue <LL> q[N];
16 multiset <LINE> s;
17 multiset <LINE> :: iterator it;
18 inline V found1 (I x,I y) {
19 to[++tot] = y,nxt[tot] = head[x],head[x] = tot;
20 }
21 V found2 (I id) {
22 while (!s.empty ()) {
23 it = s.lower_bound (line[id]);
24 if (it == s.end ()) break;
25 LINE tmp = *it;
26 if (tmp.l >= line[id].l && tmp.r <= line[id].r)
27 s.erase (it), found1 (id,tmp.id), found2 (tmp.id);
28 else break;
29 }
30 }
31 V dfs (I x) {
32 I s(m + 1);
33 for (I i(head[x]),y(to[i]); i ;i = nxt[i],y = to[i]) {
34 dfs (y); if (q[y].size () > q[s].size ()) s = y;
35 }
36 swap (q[x],q[s]);
37 for (I i(head[x]),y(to[i]); i ;i = nxt[i],y = to[i]) if (y != s) {
38 I tmp (q[y].size ());
39 for (I i(1);i <= tmp; ++ i) sup[i] = q[x].top (), q[x].pop ();
40 for (I i(1);i <= tmp; ++ i) sup[i] += q[y].top (), q[y].pop ();
41 for (I i(1);i <= tmp; ++ i) q[x].push (sup[i]);
42 }
43 q[x].push (val[x]);
44 }
45 signed main () {
46 scanf ("%d%d",&n,&m);
47 for (I i(1);i <= m; ++ i) {
48 scanf ("%d%d%d",&line[i].l,&line[i].r,&val[i]);
49 line[i].id = i, s.insert (line[i]);
50 }
51 line[0] = (LINE) {0,1,n}; found2 (0); dfs (0); I tmp (q[0].size ()); memset (sup,0,sizeof sup);
52 for (I i(1);i <= tmp; ++ i) sup[i] = q[0].top (), q[0].pop ();
53 for (I i(1);i <= m; ++ i) sup[i] += sup[i - 1], printf ("%lld ",sup[i]);
54 }
View Code
DP优化，问题提示，最优问题可以转化为差分进行维护，最
优性问题整体合并降低复杂度。
标签：tmp,NOIP,31,SGT,mid,poimod,id,模拟,define 来源： https://www.cnblogs.com/HZOI-LYM/p/15109992.html