  1 /*
  2  * Problem:  
  3  * Author:  SHJWUDP
  4  * Created Time:  2015/6/11 星期四 12:10:10
  5  * File Name: 233.cpp
  6  * State: 
  7  * Memo: 
  8  */
  9 #pragma comment(linker, "/STACK:1024000000,1024000000")  
 10 #include <iostream>
 11 #include <cstdio>
 12 #include <cstring>
 13 #include <algorithm>
 14 using namespace std;
 15 
 16 typedef long long int64;
 17 
 18 const int MaxA=5e4+7;
 19 
 20 struct Edge {
 21     int v, nt;
 22     Edge() {}
 23     Edge(int v, int nt):v(v), nt(nt) {}
 24 } edges[MaxA<<1];
 25 
 26 int head[MaxA], edgeNum;
 27 
 28 struct Fenwick {
 29     int n;
 30     int c[MaxA];
 31     
 32     void init(int n) {
 33         this->n=n;
 34         memset(c, 0, sizeof(c[0])*(n+3));
 35     }
 36 
 37     int lowbit(int x) {
 38         return x&(-x);
 39     }
 40 
 41     void update(int x, int d) {
 42         while(x>0) {
 43             c[x]+=d;
 44             x-=lowbit(x);
 45         }
 46     }
 47 
 48     int query(int x) {
 49         int res=0;
 50         while(x<=n) {
 51             res+=c[x];
 52             x+=lowbit(x);
 53         }
 54         return res;
 55     }
 56 } fw;
 57 
 58 int N, M, P;
 59 int oval[MaxA], val[MaxA];
 60 
 61 void init() {
 62     memset(head, -1, sizeof(head));
 63     edgeNum=0;
 64 }
 65 void addEdge(int u, int v) {
 66     edges[edgeNum]=Edge(v, head[u]); 
 67     head[u]=edgeNum++;
 68 }
 69 namespace LCT {
 70     int fa[MaxA];
 71     int dep[MaxA];
 72     int siz[MaxA];
 73     int son[MaxA];
 74     int top[MaxA];
 75     int w[MaxA];
 76     int id;
 77 
 78     int dfs1(int u, int d) {
 79         dep[u]=d; siz[u]=1; son[u]=-1;
 80         for(int i=head[u]; ~i; i=edges[i].nt) {
 81             Edge& e=edges[i];
 82             if(e.v==fa[u]) continue;
 83             fa[e.v]=u;
 84             siz[u]+=dfs1(e.v, d+1);
 85             if(son[u]==-1 || siz[son[u]]<siz[e.v]) son[u]=e.v;
 86         }
 87         return siz[u];
 88     }
 89 
 90     void dfs2(int u, int tp) {
 91         w[u]=++id; top[u]=tp; 
 92         if(~son[u]) dfs2(son[u], tp);
 93         for(int i=head[u]; ~i; i=edges[i].nt) {
 94             Edge& e=edges[i];
 95             if(e.v==fa[u] || e.v==son[u]) continue;
 96             dfs2(e.v, e.v);
 97         }
 98     }
 99 
100     void init() {
101         int root=1;
102         id=0;
103         fa[root]=-1;
104         dfs1(root, 0);
105         dfs2(root, root);
106         fw.init(N);
107         for(int i=1; i<=N; i++) {
108             fw.update(w[i]-1, -oval[i]);
109             fw.update(w[i], oval[i]);
110         }
111     }
112 
113     void update(int u, int v, int d) {
114         int f1=top[u], f2=top[v];
115         while(f1!=f2) {
116             if(dep[f1]<dep[f2]) { swap(f1, f2); swap(u, v); }
117             fw.update(w[f1]-1, -d); fw.update(w[u], d); 
118             u=fa[f1]; f1=top[u];
119         }
120         if(dep[u]>dep[v]) swap(u, v);
121         fw.update(w[u]-1, -d);    fw.update(w[v], d);
122     }
123 
124     int query(int x) {
125         return fw.query(w[x]);
126     }
127 }
128 int main() {
129 #ifndef ONLINE_JUDGE
130     freopen("in", "r", stdin);
131     //freopen("out", "w", stdout);
132 #endif
133     while(~scanf("%d%d%d", &N, &M, &P)) {
134         for(int i=1; i<=N; i++) {
135             scanf("%d", &oval[i]);
136         }
137         init();
138         for(int i=0; i<M; i++) {
139             int a, b;
140             scanf("%d%d", &a, &b);
141             addEdge(a, b);
142             addEdge(b, a);
143         }
144         LCT::init();
145         while(P--) {
146             char op[2];
147             scanf("%s", op);
148             if(op[0]=='Q') {
149                 int x;
150                 scanf("%d", &x);
151                 printf("%d\n", LCT::query(x));
152             } else {
153                 int a, b, c;
154                 scanf("%d%d%d", &a, &b, &c);
155                 if(op[0]=='D') c=-c;
156                 LCT::update(a, b, c);
157             }
158         }
159     }
160     return 0;
161 }

View Code

B、BZOJ 2243

题意：N个点的点AOV树，M次操作

操作：‘C a b c’：把节点a到节点b路径上所有点（包括a和b）都染成颜色c；

　　 ‘Q a b’：询问节点a到节点b（包括a和b）路径上的颜色段数量；

解：在A的基础上修改一下就好，需要理解树剖的分链而治

  1 /**************************************************************
  2     Problem: 2243
  3     User: shjwudp
  4     Language: C++
  5     Result: Accepted
  6     Time:4324 ms
  7     Memory:17388 kb
  8 ****************************************************************/
  9  
 10 /*
 11  * Problem:  
 12  * Author:  SHJWUDP
 13  * Created Time:  2015/6/11 星期四 19:27:57
 14  * File Name: 233.cpp
 15  * State: 
 16  * Memo: 
 17  */
 18 #include <iostream>
 19 #include <cstdio>
 20 #include <cstring>
 21 #include <algorithm>
 22  
 23 using namespace std;
 24  
 25 const int MaxA=1e5+7;
 26  
 27 struct Edge {
 28     int v, nt;
 29     Edge(){}
 30     Edge(int v, int nt):v(v), nt(nt){}
 31 } edges[MaxA<<1];
 32  
 33 int head[MaxA], edgeNum;
 34  
 35 struct SegmentTree {
 36     struct Node {
 37         int l, r;
 38         int cnt;
 39         int mark;
 40     } c[MaxA<<2];
 41     int n;
 42     int *val;
 43     int L, R, v;
 44 #define lson l, m, rt<<1
 45 #define rson m+1, r, rt<<1|1
 46 #define RUSH Node& u=c[rt]; Node& ls=c[rt<<1]; Node& rs=c[rt<<1|1];
 47  
 48     void pushUp(int rt) {
 49         RUSH
 50         u.cnt=ls.cnt+rs.cnt-(ls.r==rs.l?1:0);
 51         u.l=ls.l;
 52         u.r=rs.r;
 53     }
 54  
 55     void pushDown(int rt) {
 56         RUSH
 57         if(~u.mark) {
 58             ls.mark=rs.mark=ls.l=ls.r=rs.l=rs.r=u.mark;
 59             ls.cnt=rs.cnt=1;
 60             u.mark=-1;
 61         }
 62     }
 63  
 64     void doBuild(int l, int r, int rt) {
 65         c[rt].mark=-1;
 66         if(l==r) {
 67             c[rt].l=c[rt].r=val[l];
 68             c[rt].cnt=1;
 69         } else {
 70             int m=(l+r)>>1;
 71             doBuild(lson);
 72             doBuild(rson);
 73             pushUp(rt);
 74         }
 75     }
 76  
 77     void build(int n, int *val) {
 78         this->n=n; this->val=val;
 79         doBuild(1, n, 1);
 80     }
 81  
 82     void doColor(int l, int r, int rt) {
 83         if(L<=l && r<=R) {
 84             c[rt].l=c[rt].r=c[rt].mark=v;
 85             c[rt].cnt=1;
 86         } else {
 87             pushDown(rt);
 88             int m=(l+r)>>1;
 89             if(L<=m) doColor(lson);
 90             if(m<R) doColor(rson);
 91             pushUp(rt);
 92         }
 93     }
 94  
 95     void color(int L, int R, int v) {
 96         this->L=L; this->R=R; this->v=v;
 97         doColor(1, n, 1);
 98     }
 99  
100     int doQuery(int l, int r, int rt) {
101         RUSH
102         if(L<=l && r<=R) {
103             return u.cnt;
104         } else {
105             pushDown(rt);
106             int m=(l+r)>>1;
107             int res=0;
108             if(L<=m) res+=doQuery(lson);
109             if(m<R) res+=doQuery(rson)-(res&&ls.r==rs.l?1:0);
110             return res;
111         }
112     }
113  
114     int query(int L, int R) {
115         this->L=L; this->R=R;
116         return doQuery(1, n, 1);
117     }
118  
119     int qcolor(int p) {
120         int l=1, r=n, rt=1;
121         while(l<r) {
122             pushDown(rt);
123             int m=(l+r)>>1;
124             if(p<=m) r=m, rt=rt<<1;
125             else l=m+1, rt=rt<<1|1;
126         }
127         return c[rt].l;
128     }
129  
130 #undef lson
131 #undef rson
132 #undef RUSH
133 } st;
134  
135 int N, M;
136 int oval[MaxA], val[MaxA];  ///oval:原树上结点编号到值的映射
137                             ///val:树剖后生成的线性表上编号到值的映射
138 void init() {
139     edgeNum=0;
140     memset(head, -1, sizeof(head));
141 }
142 void addEdge(int u, int v) {
143     edges[edgeNum]=Edge(v, head[u]);
144     head[u]=edgeNum++;
145 }
146 namespace LCT {
147     int fa[MaxA];
148     int siz[MaxA];
149     int son[MaxA];
150     int dep[MaxA];
151     int top[MaxA];
152     int w[MaxA];
153     int id;
154  
155     int dfs1(int u, int d) {
156         siz[u]=1; dep[u]=d; son[u]=-1;
157         for(int i=head[u]; ~i; i=edges[i].nt) {
158             Edge& e=edges[i];
159             if(e.v==fa[u]) continue;
160             fa[e.v]=u;
161             siz[u]+=dfs1(e.v, d+1);
162             if(son[u]==-1 || siz[son[u]]<siz[e.v]) son[u]=e.v;
163         }
164         return siz[u];
165     }
166  
167     void dfs2(int u, int tp) {
168         w[u]=++id; top[u]=tp;
169         if(~son[u]) dfs2(son[u], tp);
170         for(int i=head[u]; ~i; i=edges[i].nt) {
171             Edge& e=edges[i];
172             if(e.v==fa[u] || e.v==son[u]) continue;
173             dfs2(e.v, e.v);
174         }
175     }
176  
177     void init() {
178         int root=1;
179         id=0;
180         fa[root]=-1;
181         dfs1(root, 0);
182         dfs2(root, root);
183         for(int i=1; i<=N; i++) {
184             val[w[i]]=oval[i];
185         }
186         st.build(N, val);
187     }
188  
189     int find(int u, int v) {
190         int res=0;
191         int f1=top[u], f2=top[v];
192         while(f1!=f2) {
193             if(dep[f1]<dep[f2]) { swap(f1, f2); swap(u, v); }
194             res+=st.query(w[f1], w[u])
195                +(fa[f1]!=-1&&st.qcolor(w[f1])==st.qcolor(w[fa[f1]])?-1:0);
196             u=fa[f1]; f1=top[u];
197         }
198         if(dep[u]>dep[v]) swap(u, v);
199         return res+st.query(w[u], w[v]);
200     }
201  
202     void update(int u, int v, int op) {
203         int f1=top[u], f2=top[v];
204         while(f1!=f2) {
205             if(dep[f1]<dep[f2]) { swap(f1, f2); swap(u, v); }
206             st.color(w[f1], w[u], op);
207             u=fa[f1]; f1=top[u];
208         }
209         if(dep[u]>dep[v]) swap(u, v);
210         st.color(w[u], w[v], op);
211     }
212 }
213 int main() {
214 #ifndef ONLINE_JUDGE
215     freopen("in", "r", stdin);
216     //freopen("out", "w", stdout);
217 #endif
218     while(~scanf("%d%d", &N, &M)) {
219         for(int i=1; i<=N; i++) {
220             scanf("%d", &oval[i]);
221         }
222         init();
223         for(int i=1; i<N; i++) {
224             int a, b;
225             scanf("%d%d", &a, &b);
226             addEdge(a, b);
227             addEdge(b, a);
228         }
229         LCT::init();
230         while(M--) {
231             char op[2];
232             scanf("%s", op);
233             if(op[0]=='C') {
234                 int a, b, c;
235                 scanf("%d%d%d", &a, &b, &c);
236                 LCT::update(a, b, c);
237             } else {
238                 int a, b;
239                 scanf("%d%d", &a, &b);
240                 printf("%d\n", LCT::find(a, b));
241             }
242         }
243     }
244     return 0;
245 }

View Code

C、POJ 2763

题意：N个点的AOE树，q次操作，你开始在S点

操作：‘0 u’：你要去u点，问你你走过路径上的所有边的边权和；

　　　‘1 i w’：将第i条边的边权值变为w；

解：以点代边，根不代表任何边，明白每个点代表哪条边就不会有问题

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 
  6 using namespace std;
  7 
  8 const int MaxA=1e5+7;
  9 
 10 struct Edge {
 11     int v, nt;
 12     int w;
 13     Edge(){}
 14     Edge(int v, int nt, int w):v(v), nt(nt), w(w) {}
 15 } edges[MaxA<<1];
 16 
 17 int head[MaxA], edgeNum;
 18 
 19 struct Fenwick {
 20     int n;
 21     int c[MaxA];
 22 
 23     void init(int n) {
 24         this->n=n;
 25         memset(c, 0, sizeof(c[0])*(n+3));
 26     }
 27 
 28     int lowbit(int x) {
 29         return x&-x;
 30     }
 31 
 32     void update(int x, int d) {
 33         while(x<=n) {
 34             c[x]+=d;
 35             x+=lowbit(x);
 36         }
 37     }
 38 
 39     int query(int x) {
 40         int res=0;
 41         while(x>0) {
 42             res+=c[x];
 43             x-=lowbit(x);
 44         }
 45         return res;
 46     }
 47 
 48     void change(int x, int d) {
 49         int tmp=query(x)-query(x-1);
 50         update(x, d-tmp);
 51     }
 52 } fw;
 53 
 54 int N, Q, S;
 55 int oval[MaxA];
 56 void init() {
 57     memset(head, -1, sizeof(head));
 58     edgeNum=0;
 59 }
 60 void addEdge(int u, int v, int w) {
 61     edges[edgeNum]=Edge(v, head[u], w);
 62     head[u]=edgeNum++;
 63 }
 64 namespace LCT {
 65     int fa[MaxA];
 66     int siz[MaxA];
 67     int son[MaxA];
 68     int dep[MaxA];
 69     int top[MaxA];
 70     int w[MaxA];
 71     int e2p[MaxA];    ///边到点的映射
 72     int id;
 73 
 74     int dfs1(int u, int d) {
 75         siz[u]=1; dep[u]=d; son[u]=-1;
 76         for(int i=head[u]; ~i; i=edges[i].nt) {
 77             Edge& e=edges[i];
 78             if(e.v==fa[u]) continue;
 79             e2p[(i>>1)+1]=e.v;    ///边的编号映射到点
 80             fa[e.v]=u; oval[e.v]=e.w;    ///以点代边，根节点不代表任何边
 81             siz[u]+=dfs1(e.v, d+1);
 82             if(son[u]==-1 || siz[son[u]]<siz[e.v]) son[u]=e.v;
 83         }
 84         return siz[u];
 85     }
 86 
 87     void dfs2(int u, int tp) {
 88         w[u]=++id; top[u]=tp;
 89         if(~son[u]) dfs2(son[u], tp);
 90         for(int i=head[u]; ~i; i=edges[i].nt) {
 91             Edge& e=edges[i];
 92             if(e.v==fa[u] || e.v==son[u]) continue;
 93             dfs2(e.v, e.v);
 94         }
 95     }
 96 
 97     void init() {
 98         int root=(N+1)>>1;
 99         id=0;
100         fa[root]=-1; oval[root]=0;///以点代边，根节点不代表任何边，因此权为0
101         dfs1(root, 0);
102         dfs2(root, root);
103         fw.init(N);
104         for(int i=1; i<=N; i++) {
105             fw.update(w[i], oval[i]);
106         }
107     }
108 
109     int query(int u, int v) {
110         int res=0;
111         int f1=top[u], f2=top[v];
112         while(f1!=f2) {
113             if(dep[f1]<dep[f2]) { swap(f1, f2); swap(u, v); }
114             res+=fw.query(w[u])-fw.query(w[f1]-1);
115             u=fa[f1]; f1=top[u];
116         }
117         if(dep[u]>dep[v]) swap(u, v);
118         return res+fw.query(w[v])-fw.query(w[u]);
119     }
120 
121     void update(int x, int d) {
122         fw.change(w[e2p[x]], d);
123     }
124 }
125 int main() {
126 #ifndef ONLINE_JUDGE
127     freopen("in", "r", stdin);
128     //freopen("out", "w", stdout);
129 #endif
130     while(~scanf("%d%d%d", &N, &Q, &S)) {
131         init();
132         for(int i=1; i<N; i++) {
133             int a, b, c;
134             scanf("%d%d%d", &a, &b, &c);
135             addEdge(a, b, c);
136             addEdge(b, a, c);
137         }
138         LCT::init();
139         while(Q--) {
140             int op;
141             scanf("%d", &op);
142             if(op==0) {
143                 int x;
144                 scanf("%d", &x);
145                 printf("%d\n", LCT::query(S, x));
146                 S=x;
147             } else {
148                 int a, b;
149                 scanf("%d%d", &a, &b);
150                 LCT::update(a, b);
151             }
152         }
153     }
154     return 0;
155 }

View Code

D、POJ 3237

题意：N个结点的AOE树，各种操作~

操作：‘CHANGE i v’：将第i条边的权值变为w；

　　　‘NEGATE a b’：将把节点a到节点b路径上所有边的权值取负；

　　　‘QUERY a b’：询问节点a到节点b路径上的最大边权值；

　　　‘DONE’：操作结束

解：和C题差不多，只是需要修改树剖分之后的操作

  1 /*
  2  * Problem:  
  3  * Author:  SHJWUDP
  4  * Created Time:  2015/6/11 星期四 15:57:34
  5  * File Name: 233.cpp
  6  * State: 
  7  * Memo: 
  8  */
  9 #include <iostream>
 10 #include <cstdio>
 11 #include <cstring>
 12 #include <algorithm>
 13 
 14 using namespace std;
 15 
 16 const int INF=0x7f7f7f7f;
 17 
 18 const int MaxA=1e4+7;
 19 
 20 struct Edge {
 21     int v, nt;
 22     int w;
 23     Edge() {}
 24     Edge(int v, int nt, int w):v(v), nt(nt), w(w){}
 25 } edges[MaxA<<1];
 26 
 27 int head[MaxA], edgeNum;
 28 
 29 struct SegmentTree {
 30     int n;
 31     struct Node {
 32         int mx, mi;
 33         int mark;
 34     } c[MaxA<<2];
 35     int *val;
 36     int p, v;
 37     int L, R;
 38 #define lson l, m, rt<<1
 39 #define rson m+1, r, rt<<1|1
 40 #define RUSH Node& u=c[rt]; Node& ls=c[rt<<1]; Node& rs=c[rt<<1|1];
 41 
 42     void pushUp(int rt) {
 43         RUSH
 44         u.mx=max(ls.mx, rs.mx);
 45         u.mi=min(ls.mi, rs.mi);
 46     }
 47 
 48     void pushDown(int rt) {
 49         RUSH
 50         if(u.mark) {
 51             ls.mx=-ls.mx; ls.mi=-ls.mi; swap(ls.mx, ls.mi);
 52             ls.mark^=1;
 53             rs.mx=-rs.mx; rs.mi=-rs.mi; swap(rs.mx, rs.mi);
 54             rs.mark^=1;
 55             u.mark=0;
 56         }
 57     }
 58 
 59     void doBuild(int l, int r, int rt) {
 60         c[rt].mark=0;
 61         if(l==r) {
 62             c[rt].mx=c[rt].mi=val[l];
 63         } else {
 64             int m=(l+r)>>1;
 65             doBuild(lson);
 66             doBuild(rson);
 67             pushUp(rt);
 68         }
 69     }
 70 
 71     void build(int n, int *val) {
 72         this->n=n; this->val=val;
 73         doBuild(1, n, 1);
 74     }
 75     
 76     void doChange(int l, int r, int rt) {
 77         if(l==r) {
 78             c[rt].mx=c[rt].mi=v;
 79         } else {
 80             pushDown(rt);
 81             int m=(l+r)>>1;
 82             if(p<=m) doChange(lson);
 83             else doChange(rson);
 84             pushUp(rt);
 85         }
 86     }
 87 
 88     void change(int p, int v) {
 89         this->p=p; this->v=v;
 90         doChange(1, n, 1);
 91     }
 92 
 93 
 94     void doNegate(int l, int r, int rt) {
 95         if(L<=l && r<=R) {
 96             c[rt].mx=-c[rt].mx; c[rt].mi=-c[rt].mi;
 97             swap(c[rt].mx, c[rt].mi);
 98             c[rt].mark^=1;
 99         } else {
100             pushDown(rt);
101             int m=(l+r)>>1;
102             if(L<=m) doNegate(lson);
103             if(m<R) doNegate(rson);
104             pushUp(rt);
105         }
106     }
107 
108     void negate(int L, int R) {
109         this->L=L; this->R=R;
110         doNegate(1, n, 1);
111     }
112 
113 
114     int doQuery(int l, int r, int rt) {
115         if(L<=l && r<=R) {
116             return c[rt].mx;
117         } else {
118             pushDown(rt);
119             int m=(l+r)>>1;
120             int res=-INF;
121             if(L<=m) res=max(res, doQuery(lson));
122             if(m<R) res=max(res, doQuery(rson));
123             return res;
124         }
125     }
126 
127     int query(int L, int R) {
128         this->L=L; this->R=R;
129         return doQuery(1, n, 1);
130     }
131 #undef lson
132 #undef rson
133 #undef RUSH
134 } st;
135 
136 int N;
137 int oval[MaxA], val[MaxA];
138 void init() {
139     edgeNum=0;
140     memset(head, -1, sizeof(head));
141 }
142 void addEdge(int u, int v, int w) {
143     edges[edgeNum]=Edge(v, head[u], w);
144     head[u]=edgeNum++;
145 }
146 namespace LCT {
147     int fa[MaxA];
148     int siz[MaxA];
149     int son[MaxA];
150     int dep[MaxA];
151     int top[MaxA];
152     int w[MaxA];
153     int id;
154     int e2p[MaxA];        ///边到点的映射
155 
156     int dfs1(int u, int d) {
157         siz[u]=1; dep[u]=d; son[u]=-1;
158         for(int i=head[u]; ~i; i=edges[i].nt) {
159             Edge& e=edges[i];
160             if(e.v==fa[u]) continue;
161             e2p[(i>>1)+1]=e.v;        ///边到点的映射
162             fa[e.v]=u; oval[e.v]=e.w;    ///以点代边
163             siz[u]+=dfs1(e.v, d+1);
164             if(son[u]==-1 || siz[son[u]]<siz[e.v]) son[u]=e.v;
165         }
166         return siz[u];
167     }
168 
169     void dfs2(int u, int tp) {
170         w[u]=++id; top[u]=tp;
171         if(~son[u]) dfs2(son[u], tp);
172         for(int i=head[u]; ~i; i=edges[i].nt) {
173             Edge& e=edges[i];
174             if(e.v==fa[u] || e.v==son[u]) continue;
175             dfs2(e.v, e.v);
176         }
177     }
178 
179     void init() {
180         int root=1;
181         id=0; oval[root]=-INF; ///以点代边，根结点不代表任何边
182         fa[root]=-1;
183         dfs1(root, 0);
184         dfs2(root, root);
185         for(int i=1; i<=N; i++) {
186             val[w[i]]=oval[i];
187         }
188         st.build(N, val);
189     }
190 
191     void change(int p, int v) {
192         st.change(w[e2p[p]], v);
193     }
194 
195     int find(int u, int v, int op) {
196         int res=-INF;
197         int f1=top[u], f2=top[v];
198         while(f1!=f2) {
199             if(dep[f1]<dep[f2]) { swap(f1, f2); swap(u, v); }
200             if(op==0) res=max(res, st.query(w[f1], w[u]));
201             else st.negate(w[f1], w[u]);
202             u=fa[f1]; f1=top[u];
203         }
204         if(u==v) return res;
205         if(dep[u]>dep[v]) swap(u, v);
206         if(op==0) res=max(res, st.query(w[u]+1, w[v]));
207         else st.negate(w[u]+1, w[v]);
208         return res;
209     }
210 }
211 int main() {
212 #ifndef ONLINE_JUDGE
213     freopen("in", "r", stdin);
214     //freopen("out", "w", stdout);
215 #endif
216     int T;
217     scanf("%d", &T);
218     while(T--) {
219         scanf("%d", &N);
220         init();
221         for(int i=1; i<N; i++) {
222             int a, b, c;
223             scanf("%d%d%d", &a, &b, &c);
224             addEdge(a, b, c);
225             addEdge(b, a, c);
226         }
227 
228         LCT::init();
229         char op[7];
230         while(scanf("%s", op), op[0]!='D') {
231             int a, b;
232             scanf("%d%d", &a, &b);
233             switch(op[0]) {
234                 case 'C': LCT::change(a, b);
235                     break;
236                 case 'N': LCT::find(a, b, 1);
237                     break;
238                 default: printf("%d\n", LCT::find(a, b, 0));
239             }
240         }
241     }
242     return 0;
243 }

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posted on 2015-06-17 22:06 shjwudp 阅读(302) 评论(0) 编辑收藏举报

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