zoukankan      html  css  js  c++  java
  • SPOJ 375 Query on a tree

    SPOJ_375

        树链剖分的处女作终于AC啦,O(∩_∩)O哈哈~

        思想是从这篇博客里面学来的:http://blog.sina.com.cn/s/blog_6974c8b20100zc61.html,上面讲得通俗易懂。只不过我在预处理的时候是用他说的避免暴栈的bfs方式来做的。

    #include<stdio.h>
    #include<string.h>
    #define MAXD 10010
    #define MAXM 20010
    #define NINF 0xc3c3c3c3
    int N, cnt, first[MAXD], e, next[MAXM], v[MAXM], fa[MAXD], dep[MAXD], size[MAXD], son[MAXD], top[MAXD], w[MAXD], max[4 * MAXD];
    int q[MAXD];
    struct Edge
    {
        int x, y, z;
    }edge[MAXD];
    int Max(int x, int y)
    {
        return x > y ? x : y;
    }
    void Swap(int &x, int &y)
    {
        int t;
        t = x, x = y, y = t;
    }
    void add(int x, int y)
    {
        v[e] = y;
        next[e] = first[x], first[x] = e ++;
    }
    void prepare()
    {
        int i, j, k, x, y, rear = 0;
        q[rear ++] = 1;
        dep[1] = 1, fa[1] = 0, son[1] = 0;
        for(i = 0; i < rear; i ++)
        {
            x = q[i];
            for(j = first[x]; j != -1; j = next[j])
                if(v[j] != fa[x])
                {
                    fa[v[j]] = x, dep[v[j]] = dep[x] + 1;
                    q[rear ++] = v[j];
                }
        }
        size[0] = 0;
        for(i = rear - 1; i >= 0; i --)
        {
            x = q[i];
            size[x] = 1, son[x] = 0;
            for(j = first[x]; j != -1; j = next[j])
                if(v[j] != fa[x])
                {
                    size[x] += size[v[j]];
                    if(size[v[j]] > size[son[x]])
                        son[x] = v[j];
                }
        }
        cnt = 0;
        memset(top, 0, sizeof(top));
        for(i = 0; i < rear; i ++)
        {
            x = q[i];
            if(!top[x])
            {
                for(y = x; y != 0; y = son[y])
                {
                    top[y] = x;
                    w[y] = ++ cnt;
                }
            }
        }
    }
    void update(int cur)
    {
        max[cur] = Max(max[cur << 1], max[cur << 1 | 1]);
    }
    void refresh(int cur, int x, int y, int k, int v)
    {
        int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
        if(x == y)
        {
            max[cur] = v;
            return ;
        }
        if(k <= mid)
            refresh(ls, x, mid, k, v);
        else
            refresh(rs, mid + 1, y, k, v);
        update(cur);
    }
    void init()
    {
        int i, x, y, z;
        e = 0;
        memset(first, -1, sizeof(first));
        scanf("%d", &N);
        for(i = 1; i < N; i ++)
        {
            scanf("%d%d%d", &edge[i].x, &edge[i].y, &edge[i].z);
            add(edge[i].x, edge[i].y), add(edge[i].y, edge[i].x);
        }
        prepare();
        memset(max, 0xc3, sizeof(max));
        for(i = 1; i < N; i ++)
        {
            if(dep[edge[i].x] > dep[edge[i].y])
                Swap(edge[i].x, edge[i].y);
            refresh(1, 1, N, w[edge[i].y], edge[i].z);
        }
    }
    void query(int cur, int x, int y, int s, int t, int &ans)
    {
        int mid = (x + y) >> 1, ls = cur << 1, rs = cur << 1 | 1;
        if(x >= s && y <= t)
        {
            ans = Max(ans, max[cur]);
            return ;
        }
        if(mid >= s)
            query(ls, x, mid, s, t, ans);
        if(mid + 1 <= t)
            query(rs, mid + 1, y, s, t, ans);
    }
    void Query(int x, int y)
    {
        int fx = top[x], fy = top[y], ans = NINF;
        while(fx != fy)
        {
            if(dep[fx] > dep[fy])
                Swap(fx, fy), Swap(x, y);
            query(1, 1, N, w[fy], w[y], ans);
            y = fa[fy], fy = top[y];
        }
        if(x != y)
        {
            if(dep[x] > dep[y])
                Swap(x, y);
            query(1, 1, N, w[son[x]], w[y], ans);
        }
        printf("%d\n", ans);
    }
    void solve()
    {
        int i, x, y;
        char op[10];
        for(;;)
        {
            scanf("%s", op);
            if(op[0] == 'D')
                break;
            scanf("%d%d", &x, &y);
            if(op[0] == 'C')
                refresh(1, 1, N, w[edge[x].y], y);
            else
                Query(x, y);
        }
    }
    int main()
    {
        int t;
        scanf("%d", &t);
        while(t --)
        {
            init();
            solve();
        }
        return 0;
    }
  • 相关阅读:
    《C语言》for语句(8)
    解决vue vue.runtime.esm.js?2b0e:619 [Vue warn]: Error in nextTick: “TypeError: Cannot convert undefine
    React中WebSocket使用以及服务端崩溃重连
    React Native 中 react-navigation 导航器的使用 [亲测可用]
    ueditor 修改内容方法报错no funtion解决方式
    nodeJs与elementUI实现多图片上传
    Vue多页面开发案例
    Vue.js Cli 3.0 多页面开发案例解析
    基于node.js 微信支付notify_url回调接收不到xml
    react-image-gallery 加入视频图片混合显示
  • 原文地址:https://www.cnblogs.com/staginner/p/2531125.html
Copyright © 2011-2022 走看看