hdu 5029 Relief grain(树链剖分+线段树)

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题目链接：hdu 5029 Relief grain

题目大意：给定一棵树，然后每次操作在uv路径上为每个节点添加一个数w，最后输出每个节点个数最多的那个数。

解题思路：因为是在树的路径上做操作，所以基本就是树链剖分了。只不过以前是用一个数组即可维护值，这题要用

一个vector数组记录。过程中用线段树维护最大值。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

const int maxn = 100010;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], s[maxn << 2], d[maxn << 2];

inline void pushup(int u) {
    int k = s[lson(u)] < s[rson(u)] ? rson(u) : lson(u);
    s[u] = s[k];
    d[u] = d[k];
}

void build (int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;
    if (l == r) {
        s[u] = 0;
        d[u] = l;
        return;
    }
    int mid = (l + r) / 2;
    build(lson(u), l, mid);
    build(rson(u), mid + 1, r);
    pushup(u);
} 

void modify(int u, int x, int v) {
    if (lc[u] == x && rc[u] == x) {
        s[u] += v;
        return;
    }

    int mid = (lc[u] + rc[u]) / 2;
    if (x <= mid)
        modify(lson(u), x, v);
    else
        modify(rson(u), x, v);
    pushup(u);
}

typedef pair<int,int> pii;
vector<pii> g[maxn];
int N, M, E, first[maxn], jump[maxn * 2], link[maxn * 2], val[maxn];
int id, idx[maxn], dep[maxn], top[maxn], far[maxn], son[maxn], cnt[maxn];

inline void add_Edge (int u, int v) {
    link[E] = v;
    jump[E] = first[u];
    first[u] = E++;
}

void dfs (int u, int pre, int d) {
    far[u] = pre;
    dep[u] = d;
    cnt[u] = 1;
    son[u] = 0;

    for (int i = first[u]; i + 1; i = jump[i]) {
        int v = link[i];
        if (v == pre)
            continue;
        dfs(v, u, d + 1);
        cnt[u] += cnt[v];
        if (cnt[son[u]] < cnt[v])
            son[u] = v;
    }
}

void dfs(int u, int rot) {
    top[u] = rot;
    idx[u] = ++id;
    if (son[u])
        dfs(son[u], rot);
    for (int i = first[u]; i + 1; i = jump[i]) {
        int v = link[i];
        if (v == far[u] || v == son[u])
            continue;
        dfs(v, v);
    }
}

void init () {
    int u, v;
    id = E = 0;
    memset(first, -1, sizeof(first));
    for (int i = 0; i <= N; i++) g[i].clear();

    for (int i = 1; i < N; i++) {
        scanf("%d%d", &u, &v);
        add_Edge(u, v);
        add_Edge(v, u);
    }
    dfs(1, 0, 0);
    dfs(1, 1);
}

inline void add(int l, int r, int x) {
    g[l].push_back(make_pair(x, 1));
    g[r+1].push_back(make_pair(x, -1));
}

void solve (int u, int v, int w) {
    int p = top[u], q = top[v];
    while (p != q) {
        if (dep[p] < dep[q]) {
            swap(p, q);
            swap(u, v);
        }
        add(idx[p], idx[u], w);
        u = far[p];
        p = top[u];
    }

    if (dep[u] > dep[v])
        swap(u, v);
    add(idx[u], idx[v], w);
}

int main () {
    while (scanf("%d%d", &N, &M) == 2 && N + M) {
        init();
        int u, v, w;
        while (M--) {
            scanf("%d%d%d", &u, &v, &w);
            solve(u, v, w);
        }

        int ans = 0;
        build(1, 0, 100000);
        for (int i = 1; i <= N; i++) {

            for (int j = 0; j < g[i].size(); j++)
                modify(1, g[i][j].first, g[i][j].second);
            val[i] = d[1];
        }

        for (int i = 1; i <= N; i++)
            printf("%d\n", val[idx[i]]);
    }
    return 0;
}

hdu 5029 Relief grain(树链剖分+线段树)

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原文地址：http://blog.csdn.net/keshuai19940722/article/details/40189611