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最新推荐文章于 2017-07-26 14:23:41 发布

原创最新推荐文章于 2017-07-26 14:23:41 发布 · 521 阅读

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本文深入探讨了树链剖分算法的应用，通过实例解析如何解决树状结构的问题。同时，对比了树链剖分与LCA算法的特点，强调了选择合适算法的重要性。实践代码展示了算法的具体实现过程。

一眼看过去是一道赤果果的树链剖分。。。但是时间卡的实在是紧，更好的做法应该是LCA啊，待我去看看，这里先练练树链剖分。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <cstdio>
using namespace std;
#define LL long long
const int N = 100005;
struct EG
{
    int u, v;
} EE[N];
struct EZ
{
    int v;
    int next;
} eg[N*2];
int cnt;
int head[N];
void add_eg(int u,int v)
{
    ++cnt;
    eg[cnt].v = v;
    eg[cnt].next = head[u];
    head[u] = cnt;
}
int dep[N], w[N], fa[N], sz[N], son[N], top[N];
void dfs_1(int v,int _fa){
    fa[v] = _fa;
    son[v] = 0;
    sz[v] = 1;
    int to;
    for(int i = head[v]; ~i; i=eg[i].next){
        to = eg[i].v;
        if(to == _fa) continue;
        dep[to] = dep[v]+1;
        dfs_1(to, v);
        if(sz[to] > sz[son[v]]){
            son[v] = to;
        }
        sz[v] += sz[to];
    }
}
int id;
void build_tree(int v,int tp){
    w[v] = ++id;
    top[v] = tp;
    if(son[v] != 0) build_tree(son[v], top[v]);
    int to;
    for(int i = head[v]; ~i; i=eg[i].next){
        to = eg[i].v;
        if(to == fa[v] || to == son[v]) continue;
        build_tree(to, to);
    }
}

LL nd_val[N], eg_val[N];
LL ans_1[N], ans_2[N];
void upd1(int x,int y,int k)
{
    while(top[x] != top[y]){
        if(dep[top[x]] < dep[top[y]]) swap(x, y);
        nd_val[w[top[x]]] += k;
        nd_val[w[x]+1] -= k;
        x = fa[top[x]];
    }
    if(dep[x] > dep[y]) swap(x, y);
    nd_val[w[x]] += k;
    nd_val[w[y]+1] -= k;
}
void upd2(int x,int y,int k)
{
    while(top[x] != top[y]){
        if(dep[top[x]] < dep[top[y]]) swap(x, y);
        eg_val[w[top[x]]] += k;
        eg_val[w[x]+1] -= k;
        x = fa[top[x]];
    }
    if(x == y) return;
    if(dep[x] > dep[y]) swap(x, y);
    eg_val[w[son[x]]] += k;
    eg_val[w[y]+1] -= k;
}
int main()
{
    int i, j, T, n, m;
    int st = 0;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        cnt = 0;
        memset(head, -1, sizeof(head));
        for(i=1; i<n; ++i){
            scanf("%d%d",&EE[i].u,&EE[i].v);
            add_eg(EE[i].u, EE[i].v);
            add_eg(EE[i].v, EE[i].u);
        }
        id = 0;
        dfs_1(1, 1);
        build_tree(1, 1);
        char op[13];
        int x, y, k;
        memset(nd_val, 0, sizeof(nd_val));
        memset(eg_val, 0, sizeof(eg_val));
        while(m--){
            scanf("%s%d%d%d",op,&x,&y,&k);
            if(op[3] == '1'){
                upd1(x, y, k);
            }else{
                upd2(x, y, k);
            }
        }
        for(i=1; i<n; ++i){
            if(dep[EE[i].u] > dep[EE[i].v]) swap(EE[i].u, EE[i].v);
        }
        LL go = 0;
        for(i=1; i<=id; ++i){
            go += nd_val[i];
            ans_1[i] = go;
        }
        go = 0;
        for(i=1; i<=id; ++i){
            go += eg_val[i];
            ans_2[i] = go;
        }
        printf("Case #%d:\n",++st);
        for(i=1; i<=n; ++i){
            if(i > 1) printf(" ");
            printf("%lld",ans_1[w[i]]);
        }
        puts("");
        for(i=1; i<n; ++i){
            if(i > 1) printf(" ");
            printf("%lld",ans_2[w[EE[i].v]]);
        }
        puts("");
    }
    return 0;
}