cf605D. Board Game(BFS 树状数组 set)

题意

题目链接

有\(n\)张牌，每张牌有四个属性\((a, b, c, d)\)，主人公有两个属性\((x, y)\)(初始时为(0, 0))

一张牌能够被使用当且仅当\(a < x, b < y\)，使用后\(x\)会变为\(c\)，\(y\)会变为\(d\)

问使用第\(n\)张牌的最小步数

Sol

直接从\((0, 0)\)开始大力BFS，那么第一次到达时就是最小的，同时记录一下前驱

现在的问题就是如何知道哪些点可以选，也就是找到所有\(a < x, b < y\)的点，可以直接树状数组+set维护

由于保证了每个元素只出现一次，因此总复杂度为\(O(nlog^2n)\)

#include<bits/stdc++.h>

#define Pair pair<int, int>

#define MP(x, y) make_pair(x, y)

#define fi first

#define se second

#define pb(x) push_back(x)

// #define int long long

#define LL long long

#define pt(x) printf("%d ", x);

#define Fin(x) {freopen(#x".in","r",stdin);}

#define Fout(x) {freopen(#x".out","w",stdout);}

using namespace std;

const int MAXN = 2e5 + 10, INF = 1e9 + 10, mod = 1e9 + 7;

const double eps = 1e-9;

void chmax(int &a, int b) {a = (a > b ? a : b);}

void chmin(int &a, int b) {a = (a < b ? a : b);}

int sqr(int x) {return x * x;}

int add(int x, int y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}

void add2(int &x, int y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}

int mul(int x, int y) {return 1ll * x * y % mod;}

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, a[MAXN], b[MAXN], c[MAXN], d[MAXN], vis[MAXN], dis[MAXN], pre[MAXN], da[MAXN], num;

#define lb(x) (x & (-x))

#define sit set<Pair>::iterator

set<Pair> T[MAXN];

void Add(int x, int v, int id) {

    for(; x <= num; x += lb(x)) T[x].insert(MP(v, id));

}

vector<int> Query(int p) {

    int x = a[p], y = b[p];

    vector<int> res;

    for(; x; x -= lb(x)) {

        set<Pair> &now = T[x];

        while(1) {

            sit it = now.lower_bound(MP(y, 0));

            if(it == now.end()) break;

            res.pb(it -> se); now.erase(it);

        }

    }

    return res;

}

void Des() {

    sort(da + 1, da + num + 1); num = unique(da + 1, da + num + 1) - da - 1;

    for(int i = 1; i <= N; i++) {

        a[i] = num - (lower_bound(da + 1, da + num + 1, a[i]) - da) + 1;

        c[i] = num - (lower_bound(da + 1, da + num + 1, c[i]) - da) + 1;

        if(i != N) Add(c[i], d[i], i);

    }

}

void print(int t) {

    printf("%d\n", dis[t]);

    for(int u = t; ~u; u = pre[u]) printf("%d ", u);

}

void BFS() {

    queue<int> q; q.push(N); pre[N] = -1; dis[N] = 1;

    while(!q.empty()) {

        int p = q.front(); q.pop();

        if(a[p] == num && !b[p]) {print(p); return ;}

        vector<int> nxt = Query(p);

        for(int i = 0, t; i < nxt.size(); i++) {

            if(vis[nxt[i]]) continue; vis[nxt[i]] = 1;

            q.push(t = nxt[i]);

            dis[t] = dis[p] + 1, pre[t] = p;

        }

    }

    puts("-1");

}

signed main() {

    N = read(); bool flag = 0;

    for(int i = 1; i <= N; i++) {

        a[i] = read(), b[i] = read(), c[i] = read(), d[i] = read();

        da[++num] = a[i];

        da[++num] = c[i];

        flag |= (!a[i] && !b[i]);

    }

    if(!flag) return puts("-1");

    Des();

    BFS();

    return 0;

}