HDU 3899

题意：给一个有N个顶点的树，每个顶点上有Ti个元素，树的每条边都有一个长度Ci。假设Cost_i为树中的每个元素到顶点i的距离之和。题目要求找出树中的一个顶点x，使得Cost_x最小。

解法：最近这类题目比较多，解法是通过两次dfs在O(n)的时间复杂度内求出每个点的Cost。具体来说，假设以顶点1为根对树进行dfs。第一次dfs计算出每个子树下面的元素个数Cunt和每个元素到顶点1的距离之和。第二次dfs再计算出每个顶点的Cost。第一次dfs比较简单，关键是第二次dfs。如果已经计算出顶点x的Cost_x，现在要计算和x相连的顶点y的Cost_y。假设一共有Total个元素，x和y相连的边的权值为Length，以顶点y为根的子树所包含的元素个数为Cunt_y。则可以看出：相对于Cost_x，有Total-Cunt_y个元素要多走Length的距离，有Cunt_y个元素要少走Length的距离。所以Cost_y=Cost_x-(Total-2*Cunt_y)*Length。所以可以通过第二遍dfs求出所有的Cost。

注意：由于n比较大，有比较极端的数据，递归深度会非常大。如果用递归写dfs，会RE，所以不能用递归。现在已经越来越多的题卡这一点了...

#include <stdio.h>#include <memory.h>const int maxn = 100005;struct Graph {    int hed[maxn], nxt[maxn*2], pnt[maxn*2], val[maxn*2];    int idx;    void init(int n) {        memset(hed + 1, -1, n * 4);        idx = 0;    }    void addedge(int x, int y, int v) {        pnt[idx] = y; val[idx] = v; nxt[idx] = hed[x]; hed[x] = idx++;        pnt[idx] = x; val[idx] = v; nxt[idx] = hed[y]; hed[y] = idx++;    }} gra;struct Node {    int r, x, p;} sta[maxn];int teams[maxn], total;__int64 cunt[maxn], cost[maxn];void dfs_1() {    int x, y, r, p, top = 0;    sta[0].x = 1; sta[0].r = 0; sta[0].p = gra.hed[1];    cunt[1] = teams[1];    cost[1] = 0;    while (true) {        p = sta[top].p;        if (p == -1) {            top--;            if (top >= 0) {                p = sta[top].p;                x = sta[top].x;                y = gra.pnt[p];                cunt[x] += cunt[y];                cost[x] += cunt[y] * gra.val[p] + cost[y];                sta[top].p = gra.nxt[p];            } else {                break;            }        } else {            x = sta[top].x;            r = sta[top].r;            y = gra.pnt[p];            if (y != r) {                ++top;                cunt[y] = teams[y];                cost[y] = 0;                sta[top].r = x;                sta[top].x = y;                sta[top].p = gra.hed[y];            } else {                sta[top].p = gra.nxt[p];            }        }    }}void dfs_2() {    int x, y, r, p, top = 0;    sta[0].x = 1; sta[0].r = 0; sta[0].p = gra.hed[1];    while (true) {        p = sta[top].p;        if (p == -1) {            top--;            if (top >= 0) {                p = sta[top].p;                sta[top].p = gra.nxt[p];            } else {                break;            }        } else {            x = sta[top].x;            r = sta[top].r;            y = gra.pnt[p];            if (y != r) {                ++top;                cost[y] = cost[x] + (total - 2 * cunt[y]) * gra.val[p];                sta[top].r = x;                sta[top].x = y;                sta[top].p = gra.hed[y];            } else {                sta[top].p = gra.nxt[p];            }        }    }}/*void dfs_1(int x, int r) {    cost[x] = 0;    cunt[x] = teams[x];    for (int p = gra.hed[x]; p != -1; p = gra.nxt[p]) {        int y = gra.pnt[p];        if (y != r) {            dfs_1(y, x);            cunt[x] += cunt[y];            cost[x] += cunt[y] * gra.val[p] + cost[y];        }    }}void dfs_2(int x, int r) {    for (int p = gra.hed[x]; p != -1; p = gra.nxt[p]) {        int y = gra.pnt[p];        if (y != r) {            cost[y] = cost[x] + (total - 2 * cunt[y]) * gra.val[p];            dfs_2(y, x);        }    }}*/int main() {    int n, x, y, v;        while (scanf("%d", &n) != EOF) {             total = 0;        for (int i = 1; i <= n; i++) {            scanf("%d", teams + i);            total += teams[i];        }        gra.init(n);        for (int i = 1; i < n; i++) {            scanf("%d %d %d", &x, &y, &v);            gra.addedge(x, y, v);        }        dfs_1();        dfs_2();        __int64 ans = cost[1];        for (int i = 2; i <= n; i++) {            if (cost[i] < ans)                ans = cost[i];        }        printf("%I64d\n", ans);    }    return 0;}