POJ2763 Housewife Wind【LCA】

题意：给一棵树，有边权。两种查询，1.当前位置到y的距离 2.修改某条边

思路：LCA + 树状数组。按从根dfs访问的顺序把图转化为链，这样就能和树状数组结合了。记录下顶点序列vs和对应的深度depth。对于每个定点记录其首次出现的位置为id。这样求lca时便可，求id[u] 到  id[v]之间depth的最小值（线段树、RMQ），对应的顶点就是lca了。树状数组方面，每次按照走过的边的顺序放入树状数组add，叶子方向正边，根方向反边。并且把这些边在树状数组中的位置记录下来，修改操作时修改。u 到  v的距离 = sum( id[u] ) + sum( id[v] ) - 2 * sum( id[lca] )。纸上画一画就知道了，重边抵消了，正好等于lca到u的距离 + lca到v的距离。可能说的不清楚，《挑战程序设计》P330开始，讲得很好。

PS：以后养成用邻接表的习惯，有的题真得比用vector快好多，贴个模板

struct Edge{    int to,next;}edge[maxn*maxn];int head[maxn],tot;void init(){    memset(head,-1,sizeof head);    tot = 0;}void add(int x,int y){    edge[tot].to = y;    edge[tot].next = head[x];    head[x] = tot++;}//for(int i = head[x]; i != -1; i = edge[i].next)

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<stdlib.h>#include<math.h>#include<vector>#include<list>#include<map>#include<stack>#include<queue>#include<algorithm>#include<numeric>#include<functional>using namespace std;typedef long long ll;typedef pair<int,int> pii;#define lson(x) 2*x#define rson(x) 2*x+1const int maxn = 200005;struct Edge{int to,cost,id,next;}edge[maxn*2];int head[maxn],tot;int rmq_n;int vs[maxn*2],depth[maxn*2],id[maxn];int es[maxn*2];int tree[maxn*4],bit_n;int root,w[maxn];struct data{int l,r,num,dep; //num -> 下标，dep -> depth最小值 }node[maxn*4];void init(int x){memset(tree,0,sizeof tree);memset(head,-1,sizeof head);tot = 0;}void addedge(int a,int b,int c,int id){edge[tot].to = b;edge[tot].cost = c;edge[tot].id = id;edge[tot].next = head[a];head[a] = tot++;edge[tot].to = a;edge[tot].cost = c;edge[tot].id = id;edge[tot].next = head[b];head[b] = tot++;}int lowbit(int k){return k & -k;}void add(int cnt,int val){while(cnt <= bit_n)       // n -> num of point{tree[cnt] += val;cnt += lowbit(cnt);}}int fid(int cnt) // 1 ~ k sum{int sum = 0;while(cnt){sum += tree[cnt];cnt -= lowbit(cnt);}return sum;}void pushup(int cnt){node[cnt].dep = min(node[lson(cnt)].dep,node[rson(cnt)].dep);if(node[lson(cnt)].dep < node[rson(cnt)].dep)node[cnt].num = node[lson(cnt)].num;elsenode[cnt].num = node[rson(cnt)].num;}void build(int x,int y, int cnt){node[cnt].l = x;node[cnt].r = y;if(x == y){node[cnt].num = vs[x];node[cnt].dep = depth[x];return;}int mid = (x+y) / 2;build(x,mid,lson(cnt));build(mid+1,y,rson(cnt));pushup(cnt);}pii fidd(int x,int y,int cnt){pii a,b;a.second = 0x3f3f3f3f;b.second = 0x3f3f3f3f;if(x == node[cnt].l && y == node[cnt].r){return make_pair(node[cnt].num,node[cnt].dep);}int fa = 2*cnt;if(x <= node[fa].r){if(y <= node[fa].r)a = fidd(x,y,fa);elsea = fidd(x,node[fa].r,fa);}fa++;if(y >= node[fa].l){if(x >= node[fa].l)b = fidd(x,y,fa);elseb = fidd(node[fa].l,y,fa);}if(a.second < b.second)return a;elsereturn b;}void dfs(int v,int p,int d,int &k){id[v] = k;vs[k] = v;depth[k++] = d;for(int i = head[v]; i != -1; i = edge[i].next){Edge &e = edge[i];if(e.to != p){add(k,e.cost);es[e.id*2] = k;dfs(e.to,v,d+1,k);vs[k] = v;depth[k++] = d;add(k,-e.cost);es[e.id*2+1] = k;}}}void yu(int x){rmq_n = 2*x-1;root = (x+1) / 2;bit_n = 2 *(x-1); //走过几条边 int k = 1;dfs(root,-1,0,k);build(1,rmq_n,1);}int main(void){int n,m,x;while(scanf("%d%d%d",&n,&m,&x)!=EOF){init(n);for(int i = 1; i < n; i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);addedge(a,b,c,i);w[i] = c;}yu(n);while(m--){int op,a,b,y,gong;scanf("%d",&op);if(op == 0){scanf("%d",&y);pii f = fidd(min(id[x],id[y]),max(id[x],id[y]),1);gong = f.first;printf("%d\n",fid(id[x])+fid(id[y])-2*fid(id[gong]));x = y;}else{scanf("%d%d",&a,&b);add(es[a*2],b-w[a]);add(es[a*2+1],w[a]-b);w[a] = b;}}}return 0;}