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  • 2019牛客暑期多校训练营(第八场)- Explorer

    LCT

    和魔法森林类似,先排序一个关键字,用LCT动态维护另一个关键字。

    这里把r从大到小排序,然后动态加边。

    因为r的顺序是确定的,每次加入新边以后,r肯定是看加入的这条边的,所以我们考虑l就好了。

    对于l,我们要求能够通过的数量,肯定是要保留合法的大范围,因为小范围包含在了大范围里。

    之后就和魔法森林一样维护1~n的链。

    #include <bits/stdc++.h>
    #define INF 0x3f3f3f3f
    #define full(a, b) memset(a, b, sizeof a)
    #define __fastIn ios::sync_with_stdio(false), cin.tie(0)
    #define pb push_back
    using namespace std;
    typedef long long LL;
    inline int lowbit(int x){ return x & (-x); }
    inline int read(){
        int ret = 0, w = 0; char ch = 0;
        while(!isdigit(ch)){
            w |= ch == '-', ch = getchar();
        }
        while(isdigit(ch)){
            ret = (ret << 3) + (ret << 1) + (ch ^ 48);
            ch = getchar();
        }
        return w ? -ret : ret;
    }
    template <typename A>
    inline A __lcm(A a, A b){ return a / __gcd(a, b) * b; }
    template <typename A, typename B, typename C>
    inline A fpow(A x, B p, C lyd){
        A ans = 1;
        for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
        return ans;
    }
    const int N = 200005;
    int n, m, tot, ch[N][2], fa[N], rev[N], mx[N], id[N], st[N], w[N];
    struct Edge{
        int u, v, l, r;
        bool operator < (const Edge &rhs) const {
            return r > rhs.r;
        }
    }e[N];
    
    int build(int val){
        ++ tot;
        w[tot] = mx[tot] = val, id[tot] = tot;
        ch[tot][0] = ch[tot][1] = rev[tot] = fa[tot] = 0;
        return tot;
    }
    
    bool isRoot(int x){
        return ch[fa[x]][0] != x && ch[fa[x]][1] != x;
    }
    
    void push_up(int rt){
        int l = ch[rt][0], r = ch[rt][1];
        id[rt] = rt, mx[rt] = w[rt];
        if(mx[l] > mx[rt]) mx[rt] = mx[l], id[rt] = id[l];
        if(mx[r] > mx[rt]) mx[rt] = mx[r], id[rt] = id[r];
    }
    
    void reverse(int x){
        rev[x] ^= 1;
        swap(ch[x][0], ch[x][1]);
    }
    
    void push_down(int x){
        if(rev[x]){
            int l = ch[x][0], r = ch[x][1];
            if(l) reverse(l);
            if(r) reverse(r);
            rev[x] ^= 1;
        }
    }
    
    void rotate(int x){
        int y = fa[x], z = fa[y], p = (ch[y][1] == x) ^ 1;
        ch[y][p ^ 1] = ch[x][p], fa[ch[x][p]] = y;
        if(!isRoot(y)) ch[z][ch[z][1] == y] = x;
        fa[x] = z, fa[y] = x, ch[x][p] = y;
        push_up(y), push_up(x);
    }
    
    void splay(int x){
        int pos = 0; st[++ pos] = x;
        for(int i = x; !isRoot(i); i = fa[i]) st[++ pos] = fa[i];
        while(pos) push_down(st[pos --]);
        while(!isRoot(x)){
            int y = fa[x], z = fa[y];
            if(!isRoot(y)){
                (ch[y][0] == x) ^ (ch[z][0] == y) ? rotate(x) : rotate(y);
            }
            rotate(x);
        }
        push_up(x);
    }
    
    void access(int x){
        for(int p = 0; x; p = x, x = fa[x])
            splay(x), ch[x][1] = p, push_up(x);
    }
    
    void makeRoot(int x){
        access(x), splay(x), reverse(x);
    }
    
    void link(int x, int y){
        makeRoot(x), fa[x] = y;
    }
    
    int findRoot(int x){
        access(x), splay(x);
        while(ch[x][0]) push_down(x), x = ch[x][0];
        splay(x);
        return x;
    }
    
    void split(int x, int y){
        makeRoot(x), access(y), splay(y);
    }
    
    bool isConnect(int x, int y){
        makeRoot(x);
        return findRoot(y) == x;
    }
    
    vector<pair<int, int>> ans;
    
    int main(){
    
        n = read(), m = read();
        for(int i = 1; i <= m; i ++){
            e[i].u = read(), e[i].v = read(), e[i].l = read(), e[i].r = read();
        }
        for(int i = 1; i <= n; i ++) build(0);
        sort(e + 1, e + m + 1);
        for(int i = 1; i <= m; i ++){
            int u = e[i].u, v = e[i].v, t = build(e[i].l);
            if(!isConnect(u, v)){
                link(u, t), link(t, v);
            }
            else{
                split(u, v);
                if(mx[v] >= e[i].l){
                    int p = id[v];
                    splay(p);
                    fa[ch[p][0]] = 0, fa[ch[p][1]] = 0;
                    ch[p][0] = ch[p][1] = 0;
                    link(u, t), link(t, v);
                }
            }
            if(isConnect(1, n)){
                split(1, n);
                if(mx[n] <= e[i].r) ans.pb(make_pair(mx[n], e[i].r));
            }
        }
        sort(ans.begin(), ans.end());
        int res = 0;
        for(int i = 0; i < ans.size(); i ++){
            int j = i, cur = ans[i].second;
            while(j + 1 < ans.size() && ans[j + 1].first <= cur) cur = max(cur, ans[++ j].second);
            res += cur - ans[i].first + 1;
            i = j;
        }
        printf("%d
    ", res);
    
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/onionQAQ/p/11341382.html
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