• 洛谷P4069 [SDOI2016]游戏(李超线段树)


    题意

    题目链接

    Sol

    这题细节好多啊qwq。。稍不留神写出一个小bug就要调1h+。。

    思路就不多说了,把询问区间拆成两段就是李超线段树板子题了。

    关于dis的问题可以直接维护。

    // luogu-judger-enable-o2
    /*
    李超线段树板子题 
    */
    #include<bits/stdc++.h> 
    #define Pair pair<int, int>
    #define MP(x, y) make_pair(x, y)
    #define fi first
    #define se second
    #define int long long 
    #define LL long long 
    #define Fin(x) {freopen(#x".in","r",stdin);}
    #define Fout(x) {freopen(#x".out","w",stdout);}
    using namespace std;
    const int MAXN = 1e6 + 10;
    template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
    template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
    inline int read() {
        char c = getchar(); int x = 0, f = 1;
        while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
        while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
        return x * f;
    }
    int N, M, dis[MAXN], son[MAXN], fa[MAXN], top[MAXN], dep[MAXN], siz[MAXN], ID[MAXN], rev[MAXN], root, tot;
    LL val[MAXN];
    vector<Pair> v[MAXN];
    struct Line {
        int k, b;
    }s[MAXN];
    int ls[MAXN], rs[MAXN], ll[MAXN], rr[MAXN], mn[MAXN];
    void dfs1(int x, int _fa) {
        fa[x] = _fa; dep[x] = dep[_fa] + 1; siz[x] = 1;
        for(auto &to : v[x]) {
            if(to.fi == _fa) continue;
            dis[to.fi] = dis[x] + to.se;
            dfs1(to.fi, x);
            siz[x] += siz[to.fi];
            if(siz[to.fi] > siz[son[x]]) son[x] = to.fi;
        }
    }
    void dfs2(int x, int topf) {	
        top[x] = topf; ID[x] = ++ID[0]; rev[ID[0]] = x;
        if(!son[x]) return ;
        dfs2(son[x], topf);
        for(auto &to : v[x]) {
            if(top[to.fi]) continue;
            dfs2(to.fi, to.fi);
        }
    }
    int LCA(int x, int y) {
        while(top[x] ^ top[y]) {
            if(dep[top[x]] < dep[top[y]]) swap(x, y);
            x = fa[top[x]];
        }
        return dep[x] > dep[y] ? y : x;
    }
    int calc(Line s, int x) {
        return s.k * dis[rev[x]] + s.b;
    }
    double GetPoint(Line x, Line y) {
        return (double) (y.b - x.b) / (x.k - y.k);
    }
    void update(int k) {
        if(s[k].k || s[k].b) chmin(mn[k], min(calc(s[k], ll[k]), calc(s[k], rr[k])));
        chmin(mn[k], min(mn[ls[k]], mn[rs[k]]));
    
    }
    void IntAdd(int &k, int l, int r, int ql, int qr, Line seg) {
        if(!k) k = ++tot, mn[k] = val[0], ll[k] = l, rr[k] = r;
        int mid = l + r >> 1;
        if(ql <= l && r <= qr) {
            int pl = calc(s[k], l), pr = calc(s[k], r), nl = calc(seg, l), nr = calc(seg, r);
            if(!s[k].k && !s[k].b) {s[k] = seg; update(k); return ;}
            if(pl <= nl && pr <= nr) return ;//这里必须要写等号 
            if(pl > nl && pr > nr) {s[k] = seg;  update(k); return ;}
            double xx = GetPoint(s[k], seg);
            int m = dis[rev[mid]];
            if(pl > nl) {
                if(xx > m) IntAdd(rs[k], mid + 1, r, ql, qr, s[k]), s[k] = seg;
             	else IntAdd(ls[k], l, mid, ql, qr, seg);
            } else {
                if(xx > m) IntAdd(rs[k], mid + 1, r, ql, qr, seg);
                else IntAdd(ls[k], l, mid, ql, qr, s[k]), s[k] = seg;
            }
            update(k);
            return ;
        }
        if(ql <= mid) IntAdd(ls[k], l, mid, ql, qr, seg);
        if(qr  > mid) IntAdd(rs[k], mid + 1, r, ql, qr, seg);	
        update(k);
    }
    int IntMin(int k, int l, int r, int ql, int qr) {
        int ret = val[0];
        if(ql <= l && r <= qr) 
            return mn[k];
        int mid = l + r >> 1;
        if(s[k].k || s[k].b) 
            chmin(ret, min(calc(s[k], max(l, ql)), calc(s[k], min(r, qr))));
        if(ql <= mid) chmin(ret, IntMin(ls[k], l, mid, ql, qr));
        if(qr  > mid) chmin(ret, IntMin(rs[k], mid + 1, r, ql, qr));
        return ret;
    }
    void Modify(int x, int y, int k, int b) {
        while(top[x] != top[y]) {
            if(dep[top[x]] < dep[top[y]]) swap(x, y);
            IntAdd(root, 1, N, ID[top[x]], ID[x], {k, b});
            x = fa[top[x]];
        }
        if(dep[x] < dep[y]) swap(x, y);
        IntAdd(root, 1, N, ID[y], ID[x], {k, b});
    }
    void Change(int s, int t, int a, int b) {
        int lca = LCA(s, t);
        Modify(s, lca, -a, b + a * dis[s]);
        Modify(t, lca, a, -a * dis[lca] + b + a * (dis[s] - dis[lca]));
    }
    LL Query(int x, int y) {
        int ans = val[0];
        while(top[x] != top[y]) {
            if(dep[top[x]] < dep[top[y]]) swap(x, y);
            chmin(ans, IntMin(root, 1, N, ID[top[x]], ID[x]));
            x = fa[top[x]];
        }
        if(dep[x] < dep[y]) swap(x, y);
        chmin(ans, IntMin(root, 1, N, ID[y], ID[x]));
        return ans;
    }
    signed main() {
        N = read(); M = read();
        for(int i = 0; i <= N; i++) mn[i] = val[i] = 123456789123456789ll;
        for(int i = 1; i <= N - 1; i++) {
            int x = read(), y = read(), z = read(); 
            v[x].push_back(MP(y, z));
            v[y].push_back(MP(x, z));
        }
        dfs1(1, 0);
        dfs2(1, 1);
        int GG =0;
        for(int i = 1; i <= M; i++) {
        	int opt = read(), s = read(), t = read();
        	if(opt == 1) {
        		int a = read(), b = read();
                Change(s, t, a, b);
            } else {
                cout << Query(s, t) << '
    ';
            }
        }
        return 0;
    }
    
  • 相关阅读:
    LeetCode | Remove Duplicates from Sorted List
    LeetCode | Remove Duplicates from Sorted Array
    LeetCode | Merge Sorted Array
    opencv的配置、使用
    LeetCode | Merge Two Sorted Lists
    LeetCode | Minimum Depth of Binary Tree
    LeetCode | Same Tree
    LeetCode | Maximum Depth of Binary Tree
    LeetCode | Insertion Sort List
    python运算符优先级
  • 原文地址:https://www.cnblogs.com/zwfymqz/p/10355943.html
Copyright © 2020-2023  润新知