看错题目了,想成每个城市都可以买一个东西,然后在后面的某个城市卖掉,问最大收益.这个可以类似维护上升序列的方法在O(nlog^3n)的时间复杂度内搞定
这道题用到的一些方法:
1. 可以将有关的线段提取出来,然后一起处理.
2. 线段树可以维护两个方向的信息,这样就可以处理树上有序的东西.
1 /**************************************************************
2 Problem: 3999
3 User: idy002
4 Language: C++
5 Result: Accepted
6 Time:3204 ms
7 Memory:12144 kb
8 ****************************************************************/
9
10 #include <cstdio>
11 #include <cstring>
12 #include <algorithm>
13 #define max(a,b) ((a)>(b)?(a):(b))
14 #define N 50010
15 using namespace std;
16
17 typedef long long dnt;
18 struct Node {
19 dnt t, b, a[2], tag;
20 int lf, rg;
21 Node *ls, *rs;
22 void pushdown() {
23 if( tag ) {
24 ls->t += tag;
25 ls->b += tag;
26 ls->tag += tag;
27 rs->t += tag;
28 rs->b += tag;
29 rs->tag += tag;
30 tag = 0;
31 }
32 }
33 void update() {
34 t = max( ls->t, rs->t );
35 b = min( ls->b, rs->b );
36 a[0] = max( ls->a[0], rs->a[0] );
37 a[0] = max( a[0], rs->t-ls->b );
38 a[1] = max( ls->a[1], rs->a[1] );
39 a[1] = max( a[1], ls->t-rs->b );
40 }
41 void modify( int L, int R, int d ) {
42 if( L<=lf && rg<=R ) {
43 t += d;
44 b += d;
45 tag += d;
46 return;
47 }
48 pushdown();
49 int mid=(lf+rg)>>1;
50 if( L<=mid ) ls->modify( L, R, d );
51 if( R>mid ) rs->modify( L, R, d );
52 update();
53 }
54 }pool[N*3], *tail=pool, *root;
55
56 int n, m;
57 int head[N], dest[N<<1], next[N<<1], etot;
58 int aa[N], bb[N];
59 int fat[N], dep[N], siz[N], son[N], top[N], vid[N];
60 int qu[N], bg, ed;
61 Node *su[N], *sv[N];
62 int tu, tv;
63
64 void adde( int u, int v ) {
65 etot++;
66 dest[etot] = v;
67 next[etot] = head[u];
68 head[u] = etot;
69 }
70 Node *build( int lf, int rg ) {
71 Node *nd = ++tail;
72 if( lf==rg ) {
73 nd->b = nd->t = bb[lf];
74 nd->a[0] = nd->a[1] = 0;
75 nd->lf=lf, nd->rg=rg;
76 } else {
77 int mid=(lf+rg)>>1;
78 nd->ls = build( lf, mid );
79 nd->rs = build( mid+1, rg );
80 nd->lf=lf, nd->rg=rg;
81 nd->update();
82 }
83 return nd;
84 }
85 void fetch( Node *nd, int L, int R, Node *(&stk)[N], int &top ) {
86 int lf=nd->lf, rg=nd->rg;
87 if( L<=lf && rg<=R ) {
88 stk[++top] = nd;
89 return;
90 }
91 int mid=(lf+rg)>>1;
92 nd->pushdown();
93 if( R>mid ) fetch(nd->rs,L,R,stk,top);
94 if( L<=mid ) fetch(nd->ls,L,R,stk,top);
95 }
96 void build_dcp( int s ) {
97 // fat dep
98 fat[s] = 0;
99 dep[s] = 1;
100 qu[bg=ed=1] = s;
101 while( bg<=ed ) {
102 int u=qu[bg++];
103 for( int t=head[u]; t; t=next[t] ) {
104 int v=dest[t];
105 if( v==fat[u] ) continue;
106 fat[v] = u;
107 dep[v] = dep[u] + 1;
108 qu[++ed] = v;
109 }
110 }
111 // siz son
112 for( int i=ed; i>=1; i-- ) {
113 int u=qu[i], p=fat[u];
114 siz[u]++;
115 if( p ) {
116 siz[p] += siz[u];
117 if( siz[u]>siz[son[p]] ) son[p]=u;
118 }
119 }
120 // top vid
121 top[s] = s;
122 vid[s] = 1;
123 for( int i=1; i<=ed; i++ ) {
124 int u=qu[i];
125 int cur=vid[u]+1;
126 if( son[u] ) {
127 top[son[u]] = top[u];
128 vid[son[u]] = cur;
129 cur += siz[son[u]];
130 }
131 for( int t=head[u]; t; t=next[t] ) {
132 int v=dest[t];
133 if( v==fat[u] || v==son[u] ) continue;
134 top[v] = v;
135 vid[v] = cur;
136 cur += siz[v];
137 }
138 }
139 // segment
140 for( int i=1; i<=n; i++ )
141 bb[vid[i]] = aa[i];
142 root = build( 1, n );
143 }
144 int lca( int u, int v ) {
145 while( top[u]!=top[v] ) {
146 if( dep[top[u]]<dep[top[v]] ) swap(u,v);
147 u = fat[top[u]];
148 }
149 return dep[u]<dep[v] ? u : v;
150 }
151 dnt query( int u, int v ) {
152 if( u==v ) return 0;
153 int ca = lca(u,v);
154 tu = tv = 0;
155 while( top[u]!=top[ca] ) {
156 fetch(root,vid[top[u]],vid[u],su,tu);
157 u=fat[top[u]];
158 }
159 while( top[v]!=top[ca] ) {
160 fetch(root,vid[top[v]],vid[v],sv,tv);
161 v=fat[top[v]];
162 }
163 if( u!=ca )
164 fetch(root,vid[ca],vid[u],su,tu);
165 else
166 fetch(root,vid[ca],vid[v],sv,tv);
167 dnt curt = 0;
168 dnt rt = 0;
169 for( int i=1; i<=tv; i++ ) {
170 rt = max( rt, sv[i]->a[0] );
171 rt = max( rt, curt-sv[i]->b );
172 curt = max( curt, sv[i]->t );
173 }
174 for( int i=tu; i>=1; i-- ) {
175 rt = max( rt, su[i]->a[1] );
176 rt = max( rt, curt-su[i]->b );
177 curt = max( curt, su[i]->t );
178 }
179 return rt;
180 }
181 void modify( int u, int v, int d ) {
182 while( top[u]!=top[v] ) {
183 if( dep[top[u]]<dep[top[v]] ) swap(u,v);
184 root->modify(vid[top[u]],vid[u],d);
185 u=fat[top[u]];
186 }
187 if( dep[u]<dep[v] ) swap(u,v);
188 root->modify(vid[v],vid[u],d);
189 }
190 int main() {
191 scanf( "%d", &n );
192 for( int i=1; i<=n; i++ )
193 scanf( "%d", aa+i );
194 for( int i=1,u,v; i<n; i++ ) {
195 scanf( "%d%d", &u, &v );
196 adde( u, v );
197 adde( v, u );
198 }
199 build_dcp(1);
200 scanf( "%d", &m );
201 for( int t=1,u,v,d; t<=m; t++ ) {
202 scanf( "%d%d%d", &u, &v, &d );
203 printf( "%lld\n", query(u,v) );
204 modify(u,v,d);
205 }
206 }
时间： 2024-10-15 06:21:52