一个有趣的题目【二分答案，2-SAT，线段树优化】

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题目大意1

给定n(n≤2×104)个二元组(xi,yi)，要求从每个二元组中选择一个数构成集合S，最大化min{abs(i−j)|i,j∈S}。

分析与解

考场上最后1h想到正解…无奈以前并没有做过2-SAT，最后暴零滚粗…

首先显然二分答案ans，我们现在考虑如何验证。考虑如果我们选择了一个xi(或yi)，则∀k使得xk(yk)−xi(yi)<ans。选一个逻辑推出选另一个，于是可以用2-SAT解决这个问题。

然而2-SAT中的边数是O(n2)的，复杂度为O(n2lgW)，显然不能通过。考虑每一个节点只能影响其周围一段连续的区间，因此可以排序后用线段树将边数降到O(nlgn)，从而总的复杂度就是O(nlgnlgW)，就可以通过此题。

My_Code

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int MAXN = 130005;struct TwoSat {    struct node {        int to, next;    } edge[MAXN*120];    int n, head[MAXN*4], top;    inline int Not(int i)    { return i <= n ? n+i : i-n; }    void push(int i, int j)    { ++top, edge[top] = (node){j, head[i]}, head[i] = top; }    int dfn[MAXN*10], low[MAXN*10], stk[MAXN*10], stk_top, instk[MAXN*10];    int gp[MAXN*10], gp_top, dfn_top;    void init(int _n)    {        n = _n;        memset(dfn, 0, sizeof dfn); memset(head, 0, sizeof head);        memset(instk, 0, sizeof instk); memset(gp, 0, sizeof gp);        gp_top = dfn_top = stk_top = top = 0;    }    void tarjan(int nd)    {        dfn[nd] = low[nd] = ++dfn_top, stk[++stk_top] = nd, instk[nd] = 1;        for (int i = head[nd]; i; i = edge[i].next) {            int to = edge[i].to;            if (!dfn[to]) tarjan(to), low[nd] = min(low[nd], low[to]);            else if (instk[to]) low[nd] = min(low[nd], dfn[to]);        }        if (dfn[nd] == low[nd]) {            int now; ++gp_top;            do {                now = stk[stk_top--], gp[now] = gp_top, instk[now] = 0;            } while (now != nd);        }    }    bool work()    {        for (int i = 1; i <= n*2; i++)            if (!dfn[i])                tarjan(i);        for (int i = 1; i <= n; i++)            if (gp[i] == gp[i+n])                return false;        return true;    }} TSet;struct pr {    long long x;    int from;    friend bool operator < (const pr &a, const pr &b)    {        return a.x < b.x;    }} flag[MAXN];int n, flg = 0;int tree[(1<<18)+1], root = 0, N = 1<<17;bool judge(long long k){    TSet.init(n);    for (int i = 1; i <= n*2; i++) TSet.push(n*2+i+N-1, TSet.Not(flag[i].from));    for (int i = N-1; i >= 1; i--) {        if (i*2 <= N-1+n*2) TSet.push(n*2+i, n*2+i*2);         if (i*2+1 <= N-1+n*2) TSet.push(n*2+i, n*2+i*2+1);    }    for (int i = 1; i <= n*2; i++) {        int l = i+1, r = n*2;        if (l <= r) {            while (l <= r) {                int mid = (l+r)>>1;                if (flag[mid].x-flag[i].x < k) l = mid+1;                else r = mid-1;            }            int j, k;            if (i+1 <= l-1) {                for (j = i+1+N-1, k = l-1+N-1; j < k; j >>= 1, k >>= 1) {                    if (j&1) TSet.push(flag[i].from, n*2+j), j++;                    if (!(k&1)) TSet.push(flag[i].from, n*2+k), k--;                }                if (j == k) TSet.push(flag[i].from, n*2+j);            }        }        l = 1, r = i-1;        if (l <= r) {            while (l <= r) {                int mid = (l+r)>>1;                if (flag[i].x-flag[mid].x < k) r = mid-1;                else l = mid+1;            }             int j, k;            if (r+1 <= i-1) {                for (j = r+1+N-1, k = i-1+N-1; j < k; j >>= 1, k >>= 1) {                    if (j&1) TSet.push(flag[i].from, n*2+j), j++;                    if (!(k&1)) TSet.push(flag[i].from, n*2+k), k--;                }                if (j == k) TSet.push(flag[i].from, n*2+j);            }        }    }    return TSet.work();}int main(){    freopen("flag.in","r",stdin);    freopen("flag.out","w",stdout);    scanf("%d", &n);    for (int i = 1; i <= n; i++) {        long long a, b;        scanf("%lld%lld", &a, &b);        flag[++flg].x = a, flag[flg].from = i;        flag[++flg].x = b, flag[flg].from = i+n;    }    sort(flag+1, flag+n*2+1);    long long l = 0, r = 1e9;    while (l <= r) {        long long mid = (l+r)>>1;        if (judge(mid)) l = mid+1;        else r = mid-1;    }    cout << l-1 << endl;    return 0;}

出题人，VW ↩

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