|poj 3728|LCA|The merchant

poj 3728

维护6个数组
pre[i][j] i的第2^j个祖先
deep[i] i的深度
up[i][j] i到i的第2^j个祖先的最优解
down[i][j] i的第2^j个祖先到i的最优解
dmax[i][j] i到i的第2^j个祖先路径上的最大值
dmin[i][j] i到i的第2^j个祖先路径上的最小值
然后可以通过倍增维护以上数组，期中包含一些dp思想，此题非常好，作为一个LCA的跳板

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<cstdlib>#define ms(i, j) memset(i, j, sizeof i)#define FN2 "poj3728" using namespace std;const int ZINF2 = 100000000, MAXN = 50000 + 5, logs = 16;int n, wi[MAXN], pre[MAXN][logs+1], deep[MAXN], up[MAXN][logs+1], down[MAXN][logs+1], dmax[MAXN][logs+1], dmin[MAXN][logs+1]; /*维护6个数组     pre[i][j] i的第2^j个祖先    deep[i] i的深度    up[i][j] i到i的第2^j个祖先的最优解    down[i][j] i的第2^j个祖先到i的最优解    dmax[i][j] i到i的第2^j个祖先路径上的最大值    dmin[i][j] i到i的第2^j个祖先路径上的最小值 */vector<int> G[MAXN];void dfs(int x, int p){    pre[x][0] = p;    deep[x] = deep[p] + 1;    dmax[x][0] = max(wi[x], wi[p]);    dmin[x][0] = min(wi[x], wi[p]);    up[x][0] = max(0, wi[p]-wi[x]);    down[x][0] = max(0, wi[x]-wi[p]);    for (int i=1;i<=logs;i++)    {        int pa = pre[x][i-1];        pre[x][i] = pre[pa][i-1];        dmax[x][i] = max(dmax[pa][i-1], dmax[x][i-1]);        dmin[x][i] = min(dmin[pa][i-1], dmin[x][i-1]);        up[x][i] = max(up[pa][i-1], up[x][i-1]);//合并         up[x][i] = max(up[x][i], dmax[pa][i-1]-dmin[x][i-1]);         down[x][i] = max(down[pa][i-1], down[x][i-1]);//合并        down[x][i] = max(down[x][i], dmax[x][i-1]-dmin[pa][i-1]);     }    for (int i=0;i<G[x].size();i++)    {        int v = G[x][i];        if (p!=v)        {            dfs(v, x);        }    }}int lca(int a, int b){    if (deep[a]>deep[b]) swap(a, b);    for (int i=logs;i>=0;i--) if (deep[pre[b][i]]>=deep[a]) b = pre[b][i];    if (a==b) return a;    for (int i=logs;i>=0;i--) if (pre[a][i]!=pre[b][i]) a = pre[a][i], b = pre[b][i];    return pre[a][0];}void climb(int x, int y, bool UP, int &ans, int &temp){    temp = wi[x];    int del = deep[x] - deep[y];    for (int i=logs;i>=0;i--)    {        if ((del>>i) & 1)        {            if (UP)            {                ans = max(ans, dmax[x][i] - temp);//有可能不在up里                temp = min(temp, dmin[x][i]);                ans = max(ans, up[x][i]);            } else             {                ans = max(ans, temp - dmin[x][i]);                temp = max(temp, dmax[x][i]);                ans = max(ans, down[x][i]);            }            x = pre[x][i];        }    }}void init(){    for (int i=1;i<=n;i++)    {        deep[i] = 0;        for (int j=0;j<=logs;j++) pre[i][j] = 0;        scanf("%d", &wi[i]);    }    for (int i=1;i<n;i++)    {        int x, y;        scanf("%d%d", &x, &y);        G[x].push_back(y), G[y].push_back(x);    }}void solve(){    dfs(1, 0);    int q;    scanf("%d" ,&q);    for (int i=1;i<=q;i++)    {        int x, y;        scanf("%d%d", &x, &y);        int px, py, ans = 0, lc = lca(x, y);        climb(x, lc, 1, ans, px); climb(y, lc, 0, ans, py);        printf("%d\n", max(ans, py - px));    }}int main(){    #ifndef ONLINE_JUDGE    freopen(FN2".in","r",stdin);    freopen(FN2".out","w",stdout);    #endif    while (scanf("%d", &n)==1)    {        init();        solve();    }    return 0;}