POJ 2763 Housewife Wind

题意： 知道了一颗有 n 个节点的树和树上每条边的权值，对应两种操作：
【LCA_RMQ+树状数组】

      0 x        输出 当前节点到 x节点的最短距离，并移动到 x 节点位置      1 x val   把第 x 条边的权值改为 val

分析： 树上两个节点a,b的距离可以转化为

      dis[a] + dis[b] - 2*dis[lca(a,b)]      其中 dis[i] 表示 i 节点到根的距离，      由于每次修改一条边，树中在这条边下方的 dis[] 值全都会受到影响，这样每条边都对应这一段这条边的管辖区，      可以深搜保存遍历该点的时间戳，ll[i] 表示第一次遍历到该点的时间戳， rr[i] 表示回溯到该点时的时间戳，这样每次      修改边 i 的时候就可以对区间 [ ll[i], rr[i] ] 进行成段更新，成段更新的方式可以在 位置 ll[i] 上加一个权值，在位置      rr[i]+1 上减去这个权值，求和时，sum(ll[i]) 即为该点到根的距离。

#include<stdio.h>#include<string.h>#include<math.h>#define clr(x)memset(x,0,sizeof(x))#define maxn 200005struct node{    int to,next,w,xu;}e[1000000];int tot;int head[maxn];void add(int s,int t,int wi,int xu){    e[tot].xu=xu;    e[tot].w=wi;    e[tot].to=t;    e[tot].next=head[s];    head[s]=tot++;}int dp[maxn<<1][18];int x[maxn<<1];int d[maxn];int r[maxn];int v[maxn];int f[maxn];int ll[maxn];int rr[maxn];int g[maxn];int n,m;int min(int i,int j){    return d[i]<d[j]?i:j;}void makermq(int nn){    int i,j;    for(i=0;i<nn;i++)        dp[i][0]=i;    for(j=1;(1<<j)<=nn;j++)        for(i=1;i+(1<<j)-1<nn;i++)            dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);}int rmq(int l,int r){    int k=(int)(log((r-l+1)*1.0)/log(2.0));    return min(dp[l][k],dp[r-(1<<k)+1][k]);}int cnt,ti;void dfs(int u,int deep){    v[u]=1;    x[cnt]=u;    d[cnt]=deep;    r[u]=cnt++;    ll[u]=++ti;    int i,k;    for(i=head[u];i;i=e[i].next)    {        k=e[i].to;        if(!v[k])        {            g[e[i].xu]=k;            dfs(k,deep+1);            x[cnt]=u;            d[cnt++]=deep;        }    }    rr[u]=ti;}int tree[maxn];int lowbit(int x){    return (x)&(-x);}void update(int pos,int x){    while(pos<=n)    {        tree[pos]+=x;        pos+=lowbit(pos);    }}int sum(int pos){    int s=0;    while(pos>0)    {        s+=tree[pos];        pos-=lowbit(pos);    }    return s;}int edge[maxn];int val[maxn];int main(){    int i,st;    while(scanf("%d%d%d",&n,&m,&st)!=EOF)    {        int a,b,c;        clr(head);  clr(v);        clr(f);     clr(tree);        tot=1;        ti=-1;        for(i=1;i<n;i++)        {            scanf("%d%d%d",&a,&b,&c);            val[i]=c;            edge[i]=c;            add(a,b,c,i);            add(b,a,c,i);        }        cnt=0;        dfs(1,0);        makermq(2*n-1);        for(i=1;i<n;i++)        {            update(ll[g[i]],edge[i]);            update(rr[g[i]]+1,-edge[i]);        }        int op;        while(m--)        {            scanf("%d",&op);            if(op==1)            {                scanf("%d%d",&a,&b);                update(ll[g[a]],-val[a]);                update(rr[g[a]]+1,val[a]);                update(ll[g[a]],b);                update(rr[g[a]]+1,-b);                val[a]=b;            }            else            {                scanf("%d",&a);                int lca,d1,d2,d3;                if(r[st]<=r[a])                     lca=x[rmq(r[st],r[a])];                else lca=x[rmq(r[a],r[st])];                d1=sum(ll[st]);                d2=sum(ll[a]);                d3=sum(ll[lca]);                st=a;                printf("%d\n",d1+d2-2*d3);            }        }    }    return 0;}

这个是利用线段树维护区间和，把边尾所涉及到的时间戳区间内的值都加上对应权值，然后某个点的值就代表了它到根节点的距离！！

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 200010;struct edge{  int u, v, w, next;}edges[maxn*2], e[maxn];int E[maxn*2], H[maxn*2], I[maxn*2], L[maxn], R[maxn];int dp[maxn*2][40];int cnt, clock, dfn;int first[maxn];int a[maxn<<2];int b[maxn];int add[maxn<<2];int degree[maxn];int vis[maxn];void AddEdge(int u, int v, int w){  edges[cnt].u = u;  edges[cnt].v = v;  edges[cnt].w = w;  edges[cnt].next = first[u];  first[u] = cnt++;  edges[cnt].u = v;  edges[cnt].v = u;  edges[cnt].w = w;  edges[cnt].next = first[v];  first[v] = cnt++; }void dfs(int u, int fa, int dep){  E[++clock] = u;  H[clock] = dep;  I[u] = clock;  L[u] = ++dfn;  b[dfn] = u;  for(int i = first[u]; i != -1; i = edges[i].next)  {    int v = edges[i].v;    if(v == fa)      continue;    if(vis[v])      continue;    vis[v] = true;    dfs(v, u, dep+1);    E[++clock] = u;    H[clock] = dep;  }  R[u] = dfn;}void RMQ_init(int n){  for(int i = 1; i <= n; i++)    dp[i][0] = i;  for(int j = 1; (1<<j) <= n; j++)    for(int i = 1; i+(1<<j)-1 <= n; i++)    {      if(H[dp[i][j-1]] < H[dp[i+(1<<(j-1))][j-1]])        dp[i][j] = dp[i][j-1];      else        dp[i][j] = dp[i+(1<<(j-1))][j-1];    }}int RMQ(int l, int r){  l = I[l], r = I[r];  if(l > r)    swap(l, r);  int len = r-l+1, k = 0;  while((1<<k) <= len)    k++;  k--;  if(H[dp[l][k]] < H[dp[r-(1<<k)+1][k]])    return E[dp[l][k]];  else    return E[dp[r-(1<<k)+1][k]];}void pushdown(int rt, int l, int r){  int k = (r-l+1);  if(add[rt])  {    a[rt<<1] += add[rt]*(k-(k>>1));    a[rt<<1|1] += add[rt]*(k>>1);    add[rt<<1] += add[rt];    add[rt<<1|1] += add[rt];    add[rt] = 0;  }}void build(int l, int r, int rt){  a[rt] = 0;  add[rt] = 0;  if(l == r)    return;  int m = (l + r) >> 1;  build(l, m, rt<<1);  build(m+1, r, rt<<1|1);}void update(int x, int y, int l, int r, int rt, int num){  if(l == x && r == y)  {    a[rt] += (r-l+1)*num;    add[rt] += num;    return;  }  pushdown(rt, l, r);  int m = (l + r) >> 1;  if(y <= m)    update(x, y, l, m, rt<<1, num);  else if(x > m)    update(x, y, m+1, r, rt<<1|1, num);  else  {    update(x, m, l, m, rt<<1, num);    update(m+1, y, m+1, r, rt<<1|1, num);  }  a[rt] = a[rt<<1] + a[rt<<1|1];}int query(int x, int l, int r, int rt){  if(l == r)  {    return a[rt];  }  pushdown(rt, l, r);  int m = (l + r) >> 1;  int ans = 0;  if(x <= m)    ans = query(x, l, m, rt<<1);  else    ans = query(x, m+1, r, rt<<1|1);  a[rt] = a[rt<<1] + a[rt<<1|1];  return ans;}int main(){  int cas = 1;  int T;  //scanf("%d", &T);  int s, to, root, n, q;  while(scanf("%d %d %d", &n, &q, &s) != EOF)  {    memset(vis, 0, sizeof(vis));    memset(first, -1, sizeof(first));    memset(degree, 0, sizeof(degree));    clock = cnt = dfn = 0;    build(1, n, 1);    //for(int i = 1; i <= n; i++)    //  scanf("%d", &b[i]);    for(int i = 1; i < n; i++)    {      int u, v, w;      scanf("%d %d %d", &u, &v, &w);      e[i].u = u;      e[i].v = v;      e[i].w = w;      AddEdge(u, v, 0);      degree[v]++;    }    for(int i = 1; i <= n; i++)      if(!degree[i])      {        vis[i] = true;        dfs(i, -1, 0);        root = i;        break;      }    RMQ_init(2*n-1);    //puts("1");    for(int i = 1; i < n; i++)    {      int u = e[i].u;      int v = e[i].v;      int w = e[i].w;      //printf("***%d %d\n", L[v], R[v]);      if(L[u] < L[v])        update(L[v], R[v], 1, n, 1, w);      else        update(L[u], R[u], 1, n, 1, w);    }    while(q--)    {      int x;      scanf("%d", &x);      if(!x)      {        scanf("%d", &to);        int d1 = query(L[s], 1, n, 1);        int d2 = query(L[to], 1, n, 1);        int lca = RMQ(s, to);        int d3 = query(L[lca], 1, n, 1);        //printf("***%d %d %d\n", d1, d2, d3);        printf("%d\n", d1+d2-2*d3);        //printf("%d\n", dfn);        s = to;      }      else      {        int i, w;        scanf("%d %d", &i, &w);        int x = w - e[i].w;        e[i].w = w;        int v = e[i].v;        int u = e[i].u;        if(L[u] < L[v])          update(L[v], R[v], 1, n, 1, x);        else          update(L[u], R[u], 1, n, 1, x);      }    }  }  return 0;}

树链剖分

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;#define Del(a,b) memset(a,b,sizeof(a))const int N = 100005;int dep[N],siz[N],fa[N],id[N],son[N],val[N],top[N]; //top 最近的重链父节点int num,head[N],cnt;struct Edge{    int to,next;};Edge v[N*2];struct tree{    int x,y,val;    void read(){        scanf("%d%d%d",&x,&y,&val);    }};void add_Node(int x,int y){    v[cnt].to = y;    v[cnt].next = head[x];    head[x] = cnt++;}tree e[N];void dfs1(int u, int f, int d) {    dep[u] = d;    siz[u] = 1;    son[u] = 0;    fa[u] = f;    for(int i = head[u]; i != -1 ;i = v[i].next)    {        int to = v[i].to;        if(to != f)        {            dfs1(to,u,d+1);            siz[u] += siz[to];            if(siz[son[u]] < siz[to])                son[u] = to;        }    }}void dfs2(int u, int tp) {    top[u] = tp;    id[u] = ++num;    if (son[u]) dfs2(son[u], tp);    for(int i = head[u]; i != -1 ;i = v[i].next )    {        int to = v[i].to;        if(to == fa[u] || to == son[u])            continue;        dfs2(to,to);    }}#define lson(x) ((x<<1))#define rson(x) ((x<<1)+1)struct Tree{    int l,r,val;};Tree tree[4*N];void pushup(int x) {    tree[x].val = tree[lson(x)].val + tree[rson(x)].val;}void build(int l,int r,int v){    tree[v].l=l;    tree[v].r=r;    if(l==r){        tree[v].val = val[l];        return ;    }    int mid=(l+r)>>1;    build(l,mid,v*2);    build(mid+1,r,v*2+1);    pushup(v);}void update(int o,int v,int val)  //log(n){    if(tree[o].l==tree[o].r)    {        tree[o].val = val;        return ;    }    int mid = (tree[o].l+tree[o].r)/2;    if(v<=mid)        update(o*2,v,val);    else        update(o*2+1,v,val);    pushup(o);}int query(int o,int l, int r){    if (tree[o].l >= l && tree[o].r <= r) {        return tree[o].val;    }    int mid = (tree[o].l + tree[o].r) / 2;    if(r<=mid)        return query(o+o,l,r);    else if(l>mid)        return query(o+o+1,l,r);    else        return query(o+o,l,mid) + query(o+o+1,mid+1,r);}int Yougth(int u, int v) {    int tp1 = top[u], tp2 = top[v];    int ans = 0;    while (tp1 != tp2) {        if (dep[tp1] < dep[tp2]) {            swap(tp1, tp2);            swap(u, v);        }        ans += query(1,id[tp1], id[u]);        u = fa[tp1];        tp1 = top[u];    }    if (u == v) return ans;    if (dep[u] > dep[v]) swap(u, v);    ans += query(1,id[son[u]], id[v]);    return ans;}int main(){    int n,m,s;    while(~scanf("%d%d%d",&n,&m,&s))    {        cnt = 0;        memset(head,-1,sizeof(head));        for(int i=1;i<n;i++)        {            e[i].read();            add_Node(e[i].x,e[i].y);            add_Node(e[i].y,e[i].x);        }        num = 0;        dfs1(1,0,1);        dfs2(1,1);        for(int i=1;i<n;i++)        {            if(dep[e[i].x] < dep[e[i].y])                swap(e[i].x,e[i].y);            val[id[e[i].x]] = e[i].val;        }        build(1,num,1);        for(int i=0;i<m;i++)        {            int ok,x,y;            scanf("%d",&ok);            if(ok==0)            {                scanf("%d",&x);                printf("%d\n",Yougth(s,x));                s = x;            }            else            {                scanf("%d%d",&x,&y);                update(1,id[e[x].x],y);            }        }    }    return 0;}