因为d>=0,所以一个位置的数只会单调不降并且只会有一次穿过0.
用这个性质,我们我可在线段树中记录正数负数的个数和和,以及最大的负数以及答案.
修改操作:如果当前最大负数+d<=0,那么就直接加到懒惰标记中,否则就暴力向下传递.
因为每个节点最多被额外访问该区间负数个数次,所以所有点总共会被额外访问O(nlogn)次,在加上修改操作和询问操作的普通访问O(mlogn)次,所以时间复杂度是有保证的.
谢谢mhy12345的讲解.
1 /**************************************************************
2 Problem: 4127
3 User: idy002
4 Language: C++
5 Result: Accepted
6 Time:7872 ms
7 Memory:49256 kb
8 ****************************************************************/
9
10 #include <cstdio>
11 #include <vector>
12 #include <algorithm>
13 #define min(a,b) ((a)<(b)?(a):(b))
14 #define max(a,b) ((a)>(b)?(a):(b))
15 #define abs(a) ((a)<0?-(a):(a))
16 #define oo 0x3f3f3f3f3f3f3f3fLL
17 #define N 200010
18 //#define fprintf(...)
19 using namespace std;
20
21 typedef long long dnt;
22
23 void getn( int &v ) {
24 char ch=getchar();
25 bool neg = false;
26 while( ch!=‘-‘ && (ch<‘0‘ || ch>‘9‘) ) ch=getchar();
27 if( ch==‘-‘ ) {
28 neg = true;
29 ch = getchar();
30 }
31 v = ch-‘0‘;
32 ch = getchar();
33 while( ‘0‘<=ch && ch<=‘9‘ ) {
34 v = v*10+ch-‘0‘;
35 ch = getchar();
36 }
37 if( neg ) v=-v;
38 }
39
40 struct Node {
41 dnt ans, sum[2], cnt[2], nmx, tag;
42 Node *ls, *rs;
43 inline void init( int v ) {
44 if( v<0 ) {
45 sum[0] = v;
46 cnt[0] = 1;
47 sum[1] = 0;
48 cnt[1] = 0;
49 ans = -v;
50 tag = 0;
51 nmx = v;
52 } else {
53 sum[0] = 0;
54 cnt[0] = 0;
55 sum[1] = v;
56 cnt[1] = 1;
57 tag = 0;
58 ans = v;
59 nmx = -oo;
60 }
61 }
62 inline void update() {
63 sum[0] = ls->sum[0] + rs->sum[0];
64 sum[1] = ls->sum[1] + rs->sum[1];
65 cnt[0] = ls->cnt[0] + rs->cnt[0];
66 cnt[1] = ls->cnt[1] + rs->cnt[1];
67 ans = ls->ans + rs->ans;
68 nmx = max( ls->nmx, rs->nmx );
69 }
70 inline void pushdown() {
71 if( tag ) {
72 ls->nmx += tag;
73 rs->nmx += tag;
74 ls->sum[0] += ls->cnt[0]*tag;
75 ls->sum[1] += ls->cnt[1]*tag;
76 rs->sum[0] += rs->cnt[0]*tag;
77 rs->sum[1] += rs->cnt[1]*tag;
78 ls->ans = ls->sum[1]-ls->sum[0];
79 rs->ans = rs->sum[1]-rs->sum[0];
80 ls->tag += tag;
81 rs->tag += tag;
82 tag = 0;
83 }
84 }
85 void modify( int lf, int rg, int L, int R, int delta ) {
86 if( lf==rg ) {
87 if( cnt[0] ) {
88 if( sum[0]+delta<=0 ) {
89 sum[0]+=delta;
90 nmx = sum[0];
91 ans = -sum[0];
92 } else {
93 sum[1] = sum[0]+delta;
94 ans = sum[1];
95 cnt[1] = 1;
96 sum[0] = cnt[0] = 0;
97 nmx = -oo;
98 }
99 } else {
100 sum[1]+=delta;
101 ans = sum[1];
102 }
103 return;
104 }
105 if( L<=lf && rg<=R && nmx+delta<=0 ) {
106 nmx += delta;
107 sum[0] += cnt[0]*delta;
108 sum[1] += cnt[1]*delta;
109 ans = sum[1]-sum[0];
110 tag += delta;
111 return;
112 }
113 pushdown();
114 int mid=(lf+rg)>>1;
115 if( L<=mid ) ls->modify(lf,mid,L,R,delta);
116 if( R>mid ) rs->modify(mid+1,rg,L,R,delta);
117 update();
118 }
119 dnt query( int lf, int rg, int L, int R ) {
120 if( L<=lf && rg<=R ) return ans;
121 pushdown();
122 int mid=(lf+rg)>>1;
123 dnt rt = 0;
124 if( L<=mid ) rt+=ls->query(lf,mid,L,R);
125 if( R>mid ) rt+=rs->query(mid+1,rg,L,R);
126 update();
127 return rt;
128 }
129 }pool[N*3], *tail=pool, *root;
130
131 int n, m;
132 int aa[N], bb[N];
133 int head[N], dest[N<<1], next[N<<1], etot;
134 int fat[N], siz[N], son[N], top[N], dep[N], vid[N], qu[N], bg, ed;
135
136 void adde( int u, int v ) {
137 etot++;
138 dest[etot] = v;
139 next[etot] = head[u];
140 head[u] = etot;
141 }
142 Node *build( int lf, int rg ) {
143 Node *nd = ++tail;
144 if( lf==rg) {
145 nd->init(bb[lf]);
146 return nd;
147 }
148 int mid=(lf+rg)>>1;
149 nd->ls = build( lf, mid );
150 nd->rs = build( mid+1, rg );
151 nd->update();
152 return nd;
153 }
154 void build_dcp( int s ) {
155 // fat dep
156 fat[s] = 0;
157 dep[s] = 1;
158 qu[bg=ed=1] = s;
159 while( bg<=ed ) {
160 int u=qu[bg++];
161 for( int t=head[u]; t; t=next[t] ) {
162 int v=dest[t];
163 if( v==fat[u] ) continue;
164 fat[v] = u;
165 dep[v] = dep[u]+1;
166 qu[++ed] = v;
167 }
168 }
169 // siz son
170 for( int i=ed; i>=1; i-- ) {
171 int u=qu[i], p=fat[u];
172 siz[u]++;
173 if( p ) {
174 siz[p] += siz[u];
175 if( siz[son[p]]<siz[u] ) son[p]=u;
176 }
177 }
178 // vid top
179 vid[s] = 1;
180 top[s] = s;
181 for( int i=1; i<=ed; i++ ) {
182 int u=qu[i];
183 int cur=vid[u]+1;
184 if( son[u] ) {
185 vid[son[u]] = cur;
186 top[son[u]] = top[u];
187 cur += siz[son[u]];
188 }
189 for( int t=head[u]; t; t=next[t] ) {
190 int v=dest[t];
191 if( v==fat[u] || v==son[u] ) continue;
192 vid[v] = cur;
193 top[v] = v;
194 cur += siz[v];
195 }
196 }
197 // segment tree
198 for( int i=1; i<=n; i++ )
199 bb[vid[i]] = aa[i];
200 // for( int i=1; i<=n; i++ )
201 // fprintf( stderr, "%d ", bb[i] );
202 // fprintf( stderr, "\n" );
203 root = build( 1, n );
204 }
205 void modify( int u, int v, int delta ) {
206 while( top[u]!=top[v] ) {
207 if( dep[top[u]]<dep[top[v]] ) swap(u,v);
208 root->modify(1,n,vid[top[u]],vid[u],delta);
209 // fprintf( stderr, "modify( %d %d %d )\n", vid[top[u]], vid[u], delta );
210 u=fat[top[u]];
211 }
212 if( dep[u]<dep[v] ) swap(u,v);
213 root->modify(1,n,vid[v],vid[u],delta);
214 // fprintf( stderr, "modify( %d %d %d )\n", vid[v], vid[u], delta );
215 }
216 dnt query( int u, int v ) {
217 dnt rt = 0;
218 while( top[u]!=top[v] ) {
219 if( dep[top[u]]<dep[top[v]] ) swap(u,v);
220 rt += root->query(1,n,vid[top[u]],vid[u]);
221 // fprintf( stderr, "query( %d %d ) rt = %lld\n", vid[top[u]], vid[u], rt );
222 u=fat[top[u]];
223 }
224 if( dep[u]<dep[v] ) swap(u,v);
225 rt += root->query(1,n,vid[v],vid[u]);
226 // fprintf( stderr, "query( %d %d ) rt = %lld\n", vid[v], vid[u], rt );
227 return rt;
228 }
229 int main() {
230 getn(n); getn(m);
231 for( int i=1; i<=n; i++ )
232 getn(aa[i]);
233 for( int i=1,u,v; i<n; i++ ) {
234 getn(u); getn(v);
235 adde( u, v );
236 adde( v, u );
237 }
238 build_dcp(1);
239 for( int t=1,opt,u,v,d; t<=m; t++ ) {
240 getn(opt);
241 if( opt==1 ) {
242 getn(u);getn(v);getn(d);
243 modify( u, v, d );
244 } else {
245 getn(u); getn(v);
246 printf( "%lld\n", query(u,v) );
247 }
248 }
249 }
时间： 2024-08-29 03:13:33