HDU5002 tree

HDU5002 tree

You are given a tree with N nodes which are numbered by integers 1..N. Each node is associated with an integer as the weight. 

Your task is to deal with M operations of 4 types: 

1.Delete an edge (x, y) from the tree, and then add a new edge (a, b). We ensure that it still constitutes a tree after adding the new edge. 

2.Given two nodes a and b in the tree, change the weights of all the nodes on the path connecting node a and b (including node a and b) to a particular value x. 

3.Given two nodes a and b in the tree, increase the weights of all the nodes on the path connecting node a and b (including node a and b) by a particular value d. 

4.Given two nodes a and b in the tree, compute the second largest weight on the path connecting node a and b (including node a and b), and the number of times this weight occurs on the path. Note that here we need the strict second largest weight. For instance, the strict second largest weight of {3, 5, 2, 5, 3} is 3.

InputThe first line contains an integer T (T<=3), which means there are T test cases in the input. 

For each test case, the first line contains two integers N and M (N, M<=10^5). The second line contains N integers, and the i-th integer is the weight of the i-th node in the tree (their absolute values are not larger than 10^4). 

In next N-1 lines, there are two integers a and b (1<=a, b<=N), which means there exists an edge connecting node a and b. 

The next M lines describe the operations you have to deal with. In each line the first integer is c (1<=c<=4), which indicates the type of operation. 

If c = 1, there are four integers x, y, a, b (1<= x, y, a, b <=N) after c. 
If c = 2, there are three integers a, b, x (1<= a, b<=N, |x|<=10^4) after c. 
If c = 3, there are three integers a, b, d (1<= a, b<=N, |d|<=10^4) after c. 
If c = 4 (it is a query operation), there are two integers a, b (1<= a, b<=N) after c. 

All these parameters have the same meaning as described in problem description.OutputFor each test case, first output "Case #x:"" (x means case ID) in a separate line. 

For each query operation, output two values: the second largest weight and the number of times it occurs. If the weights of nodes on that path are all the same, just output "ALL SAME" (without quotes).Sample Input

2
3 2
1 1 2
1 2
1 3
4 1 2
4 2 3
7 7
5 3 2 1 7 3 6
1 2
1 3
3 4
3 5
4 6
4 7
4 2 6
3 4 5 -1
4 5 7
1 3 4 2 4
4 3 6
2 3 6 5
4 3 6

Sample Output

Case #1:
ALL SAME
1 2
Case #2:
3 2
1 1
3 2
ALL SAME

题解：

1：删边加边

2：路径修改

3：路径加

4：询问路径次大值及其个数

为了实现4操作，对每个节点维护最大值与次大值及个数，记为mx1,mx2,c1,c2，设x的值为v

比如往节点x加入c个值为val的数，用 solve(x,val,c) 处理

则应分四种情况维护x的信息

if(val>mx1[x])mx2[x]=mx1[x],mx1[x]=val,c2[x]=c1[x],c1[x]=c;
else if(val==mx1[x])c1[x]+=c;
else if(val>mx2[x])mx2[x]=val,c2[x]=c;
else if(val==mx2[x])c2[x]+=c;

设其左右儿子为l，r在update时只需要

清空x的信息

solve(x,v[x],1);
solve(x,mx1[l],c1[l]),solve(x,mx2[l],c2[l]);
solve(x,mx1[r],c1[r]),solve(x,mx2[r],c2[r]);

参考代码：

  1 #include
  2 #include
  3 #include
  4 #include
  5 #include
  6 #include
  7 #define inf 2000000000
  8 #define ll long long 
  9 #define N 100005
 10 using namespace std;
 11 inline int read()
 12 {
 13     int x=0,f=1;char ch=getchar();
 14     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
 15     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
 16     return x*f;
 17 }
 18 int T;
 19 int n,m,top;
 20 int q[N];
 21 int c[N][2],fa[N],v[N];
 22 int mx1[N],mx2[N],c1[N],c2[N],size[N];
 23 int ta[N],tc[N];
 24 bool rev[N];
 25 void solve(int x,int val,int c)
 26 {
 27     if(val>mx1[x])mx2[x]=mx1[x],mx1[x]=val,c2[x]=c1[x],c1[x]=c;
 28     else if(val==mx1[x])c1[x]+=c;
 29     else if(val>mx2[x])mx2[x]=val,c2[x]=c;
 30     else if(val==mx2[x])c2[x]+=c;
 31 }
 32 void update(int x)
 33 {
 34     int l=c[x][0],r=c[x][1];
 35     mx1[x]=mx2[x]=-inf;c1[x]=c2[x]=0;
 36     solve(x,v[x],1);
 37     if(l)solve(x,mx1[l],c1[l]),solve(x,mx2[l],c2[l]);
 38     if(r)solve(x,mx1[r],c1[r]),solve(x,mx2[r],c2[r]);
 39     size[x]=size[l]+size[r]+1;
 40 }
 41 void add(int y,int val)
 42 {
 43     mx1[y]+=val;v[y]+=val;
 44     if(mx2[y]!=-inf)mx2[y]+=val;
 45     ta[y]+=val;
 46 }
 47 void change(int y,int val)
 48 {
 49     mx1[y]=val;v[y]=val;c1[y]=size[y];
 50     mx2[y]=-inf;c2[y]=0;
 51     tc[y]=val;
 52     if(ta[y])ta[y]=0;
 53 }
 54 void pushdown(int x)
 55 {
 56     int l=c[x][0],r=c[x][1];
 57     if(rev[x])
 58     {
 59         rev[x]^=1;rev[l]^=1;rev[r]^=1;
 60         swap(c[x][0],c[x][1]);
 61     }
 62     if(tc[x]!=-inf)
 63     {
 64         if(l)change(l,tc[x]);
 65         if(r)change(r,tc[x]);
 66         tc[x]=-inf;
 67     }
 68     if(ta[x])
 69     {
 70         if(l)add(l,ta[x]);
 71         if(r)add(r,ta[x]);
 72         ta[x]=0;
 73     }
 74 }
 75 bool isroot(int x)
 76 {
 77     return c[fa[x]][0]!=x&&c[fa[x]][1]!=x;
 78 }
 79 void rotate(int x)
 80 {
 81     int y=fa[x],z=fa[y],l,r;
 82     if(c[y][0]==x)l=0;else l=1;r=l^1;
 83     if(!isroot(y))
 84     {
 85         if(c[z][0]==y)c[z][0]=x;else c[z][1]=x;
 86     }
 87     fa[x]=z;fa[y]=x;fa[c[x][r]]=y;
 88     c[y][l]=c[x][r];c[x][r]=y;
 89     update(y);update(x);
 90 }
 91 void splay(int x)
 92 {
 93     top=0;q[++top]=x;
 94     for(int i=x;!isroot(i);i=fa[i])
 95         q[++top]=fa[i];
 96     while(top)pushdown(q[top--]);
 97     while(!isroot(x))
 98     {
 99         int y=fa[x],z=fa[y];
100         if(!isroot(y))
101         {
102             if(c[y][0]==x^c[z][0]==y)rotate(x);
103             else rotate(y);
104         }
105         rotate(x);
106     }
107 }
108 void access(int x)
109 {
110     for(int t=0;x;t=x,x=fa[x])
111         splay(x),c[x][1]=t,update(x);
112 }
113 void makeroot(int x)
114 {
115     access(x);splay(x);rev[x]^=1;
116 }
117 void link(int x,int y)
118 {
119     makeroot(x);fa[x]=y;
120 }
121 void cut(int x,int y)
122 {
123     makeroot(x);access(y);splay(y);
124     c[y][0]=fa[x]=0;update(y);
125 }
126 void query(int x,int y)
127 {
128    makeroot(x);access(y);splay(y);
129     if(c1[y]==size[y]) puts("ALL SAME");
130     else printf("%d %d\n",mx2[y],c2[y]);
131 }
132 int main()
133 {
134     T=read();
135     for(int cas=1;cas<=T;cas++)
136     {
137         printf("Case #%d:\n",cas);
138         n=read();m=read();
139         for(int i=1;i<=n;i++)
140             v[i]=read();
141         for(int i=1;i<=n;i++)
142         {
143             mx1[i]=v[i],c1[i]=1;
144             mx2[i]=-inf,c2[i]=0;
145             size[i]=1;
146         }
147         for(int i=1;i<=n;i++)
148         {
149             fa[i]=c[i][0]=c[i][1]=0;
150             ta[i]=rev[i]=0;tc[i]=-inf;
151         }
152         for(int i=1;i)
153         {
154             int u=read(),v=read();
155             link(u,v);
156         }
157         int opt,x,y,a,b,d;
158         while(m--)
159         {
160             opt=read();
161             if(opt==1)
162             {
163                 x=read();y=read();a=read();b=read();
164                 cut(x,y);link(a,b);
165             }
166             else if(opt==2)
167             {
168                 a=read();b=read();x=read();
169                 makeroot(a);access(b);splay(b);
170                 change(b,x);
171             }
172             else if(opt==3)
173             {
174                 a=read();b=read();d=read();
175                 makeroot(a);access(b);splay(b);
176                 add(b,d);
177             }
178             else 
179             {
180                 a=read();b=read();
181                 query(a,b);
182             }
183         }
184     }
185     return 0;
186 }

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posted @ 2019-01-24 14:26 SongHL 阅读( ...) 评论( ...) 编辑 收藏