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  • FZU 2082 过路费(树链剖分 边权)题解

    题意:给出每条边权值,可以更新每条边权值,询问两个点路径的最小权值

    思路:重链剖分边权化点权,让每个儿子节点继承边权。

    插点权的时候比较边的两个节点的深度,插进儿子节点中。

    代码:

    #include<cmath>
    #include<set>
    #include<map>
    #include<queue>
    #include<cstdio>
    #include<vector>
    #include<cstring>
    #include <iostream>
    #include<algorithm>
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    const int maxn = 50000 + 5;
    const int M = 50 + 5;
    const ull seed = 131;
    const int INF = 0x3f3f3f3f;
    const int MOD = 1000000007;
    int fa[maxn];
    int top[maxn];
    int sz[maxn];
    int son[maxn];
    int deep[maxn];
    int dfn[maxn], tol;
    int fd[maxn];
    
    int n, m;
    struct Edge{
        int v, w, id, next;
    }edge[maxn << 1];
    int head[maxn], tot;
    void init(){
        memset(head, -1, sizeof(head));
        tot = tol = 0;
        memset(son, -1, sizeof(son));
    }
    void addEdge(int u, int v, int w, int id){
        edge[tot].v = v;
        edge[tot].w = w;
        edge[tot].id = id;
        edge[tot].next = head[u];
        head[u] = tot++;
    }
    void dfs1(int u, int pre, int d){
        deep[u] = d;
        fa[u] = pre;
        sz[u] = 1;
        for(int i = head[u]; i != -1; i = edge[i].next){
            int v = edge[i].v;
            if(v == pre) continue;
            dfs1(v, u, d + 1);
            sz[u] += sz[v];
            if(son[u] == -1 || sz[v] > sz[son[u]])
                son[u] = v;
        }
    }
    void dfs2(int u, int tp){
        top[u] = tp;
        dfn[u] = ++tol;
        fd[tol] = u;
        if(son[u] == -1) return;
        dfs2(son[u], tp);
        for(int i = head[u]; i != -1; i = edge[i].next){
            int v = edge[i].v;
            if(v != son[u] && v != fa[u]){
                dfs2(v, v);
            }
        }
    }
    
    ll sum[maxn << 2];
    void update(int pos, int l, int r, int v, int rt){
        if(l == r){
            sum[rt] = v;
            return;
        }
        int m = (l + r) >> 1;
        if(pos <= m)
            update(pos, l, m, v, rt << 1);
        else
            update(pos, m + 1, r, v, rt << 1 | 1);
        sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
    }
    ll query(int L, int R, int l, int r, int rt){
        if(L <= l && R >= r){
            return sum[rt];
        }
        int m = (l + r) >> 1, ans = 0;
        if(L <= m)
            ans += query(L, R, l, m, rt << 1);
        if(R > m)
            ans += query(L, R, m + 1, r, rt << 1 | 1);
        sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
        return ans;
    }
    
    ll change(int u, int v){
        ll ret = 0;
        while(top[u] != top[v]){
            if(deep[top[u]] < deep[top[v]]) swap(u, v);
            ret += query(dfn[top[u]], dfn[u], 1, n, 1);
            u = fa[top[u]];
        }
        if(u == v) return ret;  //!!!
        if(deep[u] > deep[v]) swap(u, v);
        ret += query(dfn[son[u]], dfn[v], 1, n, 1);
        return ret;
    }
    int e[maxn][3];
    int main(){
        while(~scanf("%d%d", &n, &m)){
            init();
            for(int i = 1; i <= n - 1; i++){
                int u, v, w;
                scanf("%d%d%d", &u, &v, &w);
                addEdge(u, v, w, i);
                addEdge(v, u, w, i);
                e[i][0] = u, e[i][1] = v, e[i][2] = w;
            }
            dfs1(1, 0, 0);
            dfs2(1, 1);
            for(int i = 1; i <= n - 1; i++){
                int fx = e[i][0], fy = e[i][1];
                if(deep[fx] < deep[fy]) swap(fx, fy);
                update(dfn[fx], 1, n, e[i][2], 1);
            }
            while(m--){
                int op, a, b;
                scanf("%d%d%d", &op, &a, &b);
                if(op == 0){
                    int fx = e[a][0], fy = e[a][1];
                    if(deep[fx] < deep[fy]) swap(fx, fy);
                    update(dfn[fx], 1, n, b, 1);
                }
                else{
                    printf("%lld
    ", change(a, b));
                }
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/KirinSB/p/11211513.html
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