Problem 2082 过路费
Accept: 528 Submit: 1654
Time Limit: 1000 mSec Memory Limit : 32768 KB
Problem Description
有n座城市，由n-1条路相连通，使得任意两座城市之间可达。每条路有过路费，要交过路费才能通过。每条路的过路费经常会更新，现问你，当前情况下，从城市a到城市b最少要花多少过路费。
Input
有多组样例，每组样例第一行输入两个正整数n,m(2 <= n<=50000，1<=m <= 50000),接下来n-1行，每行3个正整数a b c，(1 <= a,b <= n , a != b , 1 <= c <= 1000000000).数据保证给的路使得任意两座城市互相可达。接下来输入m行，表示m个操作，操作有两种：一. 0 a b，表示更新第a条路的过路费为b，1 <= a <= n-1 ； 二. 1 a b ， 表示询问a到b最少要花多少过路费。
Output
对于每个询问，输出一行，表示最少要花的过路费。
Sample Input
2 3
1 2 1
1 1 2
0 1 2
1 2 1
Sample Output
1
2
Source
FOJ有奖月赛-2012年4月（校赛热身赛） 题解：
直接动态树维护树上两点间的和即可。。。
注意开long long。。。
1 #include<cstdio> 2 #include<cstring> 3 #include<cstdlib> 4 #include<cmath> 5 #include<iostream> 6 #include<algorithm> 7 using namespace std; 8 #define LL long long 9 #define MAXN 50010 10 struct node 11 { 12 int left,right,val; 13 LL sum; 14 }tree[MAXN*2]; 15 int father[MAXN*2],Stack[2*MAXN],rev[2*MAXN]; 16 int read() 17 { 18 int s=0,fh=1;char ch=getchar(); 19 while(ch<'0'||ch>'9'){if(fh=='-')fh=-1;ch=getchar();} 20 while(ch>='0'&&ch<='9'){s=s*10+(ch-'0');ch=getchar();} 21 return s*fh; 22 } 23 int isroot(int x) 24 { 25 return tree[father[x]].left!=x&&tree[father[x]].right!=x; 26 } 27 void pushdown(int x) 28 { 29 int l=tree[x].left,r=tree[x].right; 30 if(rev[x]!=0) 31 { 32 rev[x]^=1;rev[l]^=1;rev[r]^=1; 33 swap(tree[x].left,tree[x].right); 34 } 35 } 36 void Pushup(int x) 37 { 38 int l=tree[x].left,r=tree[x].right; 39 tree[x].sum=tree[l].sum+tree[r].sum+tree[x].val; 40 } 41 void rotate(int x) 42 { 43 int y=father[x],z=father[y]; 44 if(!isroot(y)) 45 { 46 if(tree[z].left==y)tree[z].left=x; 47 else tree[z].right=x; 48 } 49 if(tree[y].left==x) 50 { 51 father[x]=z;father[y]=x;tree[y].left=tree[x].right;tree[x].right=y;father[tree[y].left]=y; 52 } 53 else 54 { 55 father[x]=z;father[y]=x;tree[y].right=tree[x].left;tree[x].left=y;father[tree[y].right]=y; 56 } 57 Pushup(y);Pushup(x); 58 } 59 void splay(int x) 60 { 61 int top=0,i,y,z;Stack[++top]=x; 62 for(i=x;!isroot(i);i=father[i])Stack[++top]=father[i]; 63 for(i=top;i>=1;i--)pushdown(Stack[i]); 64 while(!isroot(x)) 65 { 66 y=father[x];z=father[y]; 67 if(!isroot(y)) 68 { 69 if((tree[y].left==x)^(tree[z].left==y))rotate(x); 70 else rotate(y); 71 } 72 rotate(x); 73 } 74 } 75 void access(int x) 76 { 77 int last=0; 78 while(x!=0) 79 { 80 splay(x); 81 tree[x].right=last;Pushup(x); 82 last=x;x=father[x]; 83 } 84 } 85 void makeroot(int x) 86 { 87 access(x);splay(x);rev[x]^=1; 88 } 89 void link(int u,int v) 90 { 91 makeroot(u);father[u]=v;splay(u); 92 } 93 void cut(int u,int v) 94 { 95 makeroot(u);access(v);splay(v);father[u]=tree[v].left=0; 96 } 97 int findroot(int x) 98 { 99 access(x);splay(x); 100 while(tree[x].left!=0)x=tree[x].left; 101 return x; 102 } 103 int main() 104 { 105 int n,m,i,fh,aa,bb,cc; 106 while(scanf("%d %d",&n,&m)!=EOF) 107 { 108 for(i=0;i<=2*n;i++)tree[i].sum=tree[i].left=tree[i].right=tree[i].val=father[i]=rev[i]=0; 109 for(i=1;i<n;i++) 110 { 111 aa=read();bb=read();cc=read(); 112 tree[n+i].val=cc; 113 link(aa,n+i);link(n+i,bb); 114 } 115 for(i=1;i<=m;i++) 116 { 117 fh=read();aa=read();bb=read(); 118 if(fh==0) 119 { 120 splay(n+aa); 121 tree[n+aa].val=bb; 122 } 123 else 124 { 125 makeroot(aa);access(bb);splay(bb); 126 printf("%lld ",tree[bb].sum); 127 } 128 } 129 } 130 return 0; 131 }