• POJ2763 Housewife Wind 树链剖分 边权


    POJ2763 Housewife Wind 树链剖分 边权

    传送门:http://poj.org/problem?id=2763

    题意:

    n个点的,n-1条边,有边权

    修改单边边权

    询问 输出 当前节点到 x节点的最短距离,并移动到 x 节点位置

    题解:

    树链剖分裸题

    树链剖分就是将树分割为多条边,然后利用数据结构来维护这些链的一个技巧

    • 重儿子:父亲节点的所有儿子中子树结点数目最多( sizesiz**e 最大)的结点;
    • 轻儿子:父亲节点中除了重儿子以外的儿子;
    • 重边:父亲结点和重儿子连成的边;
    • 轻边:父亲节点和轻儿子连成的边;
    • 重链:由多条重边连接而成的路径;
    • 轻链:由多条轻边连接而成的路径;

    1.求出子树大小和每个点的重儿子,处理sz数组,son数组,fa数组和dep数组

    2.连接重链,记录dfs序,每个链的顶端节点,处理出rank数组,top数组,id数组

    3.维护链上信息

    修改边权就查询点的深度大的点,用大的点去存这条边的边权,其余的就和点权的是一样的了

    代码:

    #include <set>
    #include <map>
    #include <cmath>
    #include <cstdio>
    #include <string>
    #include <vector>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    using namespace std;
    typedef long long LL;
    typedef pair<int, int> pii;
    typedef unsigned long long uLL;
    #define ls rt<<1
    #define rs rt<<1|1
    #define lson l,mid,rt<<1
    #define rson mid+1,r,rt<<1|1
    #define bug printf("*********
    ")
    #define FIN freopen("input.txt","r",stdin);
    #define FON freopen("output.txt","w+",stdout);
    #define IO ios::sync_with_stdio(false),cin.tie(0)
    #define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]
    "
    #define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]
    "
    #define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]
    "
    const int maxn = 3e5 + 5;
    const int INF = 0x3f3f3f3f;
    
    int n, m, Q;
    int a[maxn], sz[maxn], dep[maxn], fa[maxn], top[maxn], id[maxn], son[maxn], Rank[maxn];
    int sum[maxn << 2], lazy[maxn << 2];
    
    struct EDGE {
        int u, v, nt;
    } edge[maxn << 1];
    int head[maxn], summ, cnt;
    
    void add_edge(int u, int v) {
        edge[++summ].u = u; edge[summ].v = v; edge[summ].nt = head[u]; head[u] = summ;
    }
    
    void dfs1(int u) {
        sz[u] = 1; son[u] = 0;
        for (int i = head[u]; ~i; i = edge[i].nt) {
            int v = edge[i].v;
            if (v != fa[u]) {
                fa[v] = u;
                dep[v] = dep[u] + 1;
                dfs1(v);
                sz[u] += sz[v];
                if (sz[v] > sz[son[u]]) son[u] = v;
            }
        }
    }
    
    void dfs2(int u, int tp, int x) {
        top[u] = tp; id[u] = ++cnt; Rank[cnt] = u;
        if (son[u]) dfs2(son[u], tp, 1);
        for (int i = head[u]; ~i; i = edge[i].nt) {
            int v = edge[i].v;
            if (v == son[u] || v == fa[u]) continue;
            dfs2(v, v, 2);
        }
    }
    
    void init() {
        memset(head, -1, sizeof(head));
        summ = 1; cnt = 0;
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
        for (int i = 1; i <= m; i++) {
            int u, v;
            scanf("%d %d", &u, &v);
            add_edge(u, v); add_edge(v, u);
        }
        dep[1] = 1; fa[1] = 0;
        dfs1(1);
        dfs2(1, 1, 1);
    }
    
    void pushup(int rt) {
        sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
    }
    
    void pushdown(int rt, int mid) {
        if (lazy[rt]) {
            lazy[rt << 1] += lazy[rt];
            lazy[rt << 1 | 1] += lazy[rt];
            sum[rt << 1] += lazy[rt] * (mid - mid / 2);
            sum[rt << 1 | 1] += lazy[rt] * (mid / 2);
            lazy[rt] = 0;
        }
    }
    
    void build(int l, int r, int rt) {
        lazy[rt] = 0;
        if (l == r) {
            sum[rt] = a[Rank[l]];
            return;
        }
        int mid = (l + r) >> 1;
        build(lson);
        build(rson);
        pushup(rt);
    }
    
    void update(int L, int R, int val, int l, int r, int rt) {
        if (L <= l && r <= R) {
            sum[rt] += val * (r - l + 1);
            lazy[rt] += val;
            return;
        }
        pushdown(rt, r - l + 1);
        int mid = (l + r) >> 1;
        if (L <= mid) update(L, R, val, lson);
        if (mid < R) update(L, R, val, rson);
        pushup(rt);
    }
    
    int query(int pos, int l, int r, int rt) {
        if (l == r) {
            return sum[rt];
        }
        pushdown(rt, r - l + 1);
        int mid = (l + r) >> 1;
        if (pos <= mid) return query(pos, lson);
        if (mid < pos) return query(pos, rson);
    }
    
    void change(int x, int y, int val) {
        while (top[x] != top[y]) {
            if (dep[top[x]] < dep[top[y]]) std::swap(x, y);
            update(id[top[x]], id[x], val, 1, n, 1);
            x = fa[top[x]];
        }
        if (dep[x] > dep[y]) std::swap(x, y);
        update(id[x], id[y], val, 1, n, 1);
    }
    
    int main() {
    #ifndef ONLINE_JUDGE
        FIN
    #endif
        while (~scanf("%d %d %d", &n, &m, &Q)) {
            init();
            build(1, n, 1);
            while (Q--) {
                char s[2];
                int x, y, z;
                scanf("%s", s);
                if (s[0] == 'I') {
                    scanf("%d %d %d", &x, &y, &z);
                    change(x, y, z);
                }
                if (s[0] == 'D') {
                    scanf("%d %d %d", &x, &y, &z);
                    change(x, y, -z);
                }
                if (s[0] == 'Q') {
                    scanf("%d", &x);
                    printf("%d
    ", query(id[x], 1, n, 1));
                }
            }
        }
    }
    
    每一个不曾刷题的日子 都是对生命的辜负 从弱小到强大,需要一段时间的沉淀,就是现在了 ~buerdepepeqi
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  • 原文地址:https://www.cnblogs.com/buerdepepeqi/p/11204035.html
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