HDU 4618 Palindrome Sub-Array（Manacher、二分）

题意:

给定N×M,N,M≤300的矩阵,求最大的回文正方形的边长

分析:

现场过了120,我真是一口老血,明明挺难的一个题
赛后尼玛一看,卧槽O(n5)的算法过了,n2枚举点,n枚举边长,n2判断回文
赛上想对了,可是写法很绕,绕了我很久,尼玛Manacher倍增真吃空间,hdu真抠还卡空间
我想的这个正解真的是很麻烦,−−本来没想完,被队友提示二分就会了

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p1[i][j]:=i行的以j为中心的回文半径,p2[i][j]:=i列的以j为中心的回文半径
−−被卡了空间,我重复利用了一下这个数组
至于求法,就每行来一遍Manacher,然后再每列来一遍(相当于把矩阵转置之后每行来一遍)
然后绕人的开始了−−,来看题目答案的这个正方形,插入了#是这个样的

  1234567891 #########2 #2#3#3#2#3 #########4 #2#3#3#2#5 #########6 #2#3#3#2#7 #########8 #2#3#3#2#9 #########

我们已经知道第5行每个#的纵向的回文半径长度,以及第5列每个#的横向的回文半径长度
显然通过行我们可以知道列回文长度(最小的那个),通过列我们可以知道行回文长度(最小的那个)
如何快速查询呢,−−显然区间RMQ,行列分别维护300∗2个SparseTable就好了
显然这个判断回文正方形的复杂度是O(logn)的
蓝儿我们要枚举点,枚举长度,然后在对应的2个st里查询RMQ,判断是否合法,这样复杂度是O(n3logn)
这样会T,不是我想要的,赛上我想到了这里,然后队友冷不丁冒出个二分
我说这没有单调性啊,然后立马想到奇偶分开。。。
卧槽,枚举左上角点,奇偶长度二分,然后复杂度瞬间就O(n2log2n)了,然后就可以AC了
其实之前还想到哈希判断的,但是我让他们写他们不写,因为子串太多,哈希碰撞会哭,蓝儿有这么过的
数据弱要寄刀片啊

代码:

////  Created by TaoSama on 2016-02-25//  Copyright (c) 2016 TaoSama. All rights reserved.//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 3e2 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, m, a[N][N], b[N][N];int s[N << 1], p[N][N << 1];void manacher(int *a, int *p, int n) {    s[0] = '@'; s[1] = '#';    int l = 2;    for(int i = 1; i <= n; ++i) {        s[l++] = a[i];        s[l++] = '#';    }    s[l] = 0;    int mx = 0, id;    for(int i = 1; i < l; ++i) {        if(mx > i) p[i] = min(mx - i, p[2 * id - i]);        else p[i] = 1;        while(s[i - p[i]] == s[i + p[i]]) ++p[i];        if(mx < p[i] + i) mx = p[i] + i, id = i;    }}struct SparseTable {    int n, dp[10][N << 1];    void init(int _n) {        n = _n;        for(int i = 1; (1 << i) <= n; ++i)            for(int j = 1; j + (1 << i) - 1 <= n; ++j)                dp[i][j] = min(dp[i - 1][j], dp[i - 1][j + (1 << i - 1)]);    }    int RMQ(int l, int r) {        int k = 31 - __builtin_clz(r - l + 1);        return min(dp[k][l], dp[k][r - (1 << k) + 1]);    }} hor[N << 1], ver[N << 1];bool check(int x) {    int ret = 0;    for(int i = 1; i <= 2 * n + 1; i += 2) {        for(int j = 1; j <= 2 * m + 1; j += 2) {            int right = j + 2 * x, down = i + 2 * x;            if(right > 2 * m + 1 || down > 2 * n + 1) continue;            int h = hor[i + x].RMQ(j, right), v = ver[j + x].RMQ(i, down);            int cur = min(h, v);            if(cur - 1 >= x) {                return true;            }        }    }    return false;}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    while(t--) {        scanf("%d%d", &n, &m);        for(int i = 1; i <= n; ++i)            for(int j = 1; j <= m; ++j)                scanf("%d", &a[i][j]), b[j][i] = a[i][j];        //rows        for(int i = 1; i <= n; ++i) manacher(a[i], p[i], m);        for(int i = 1; i <= 2 * m + 1; ++i) {            for(int j = 1; j <= 2 * n + 1; ++j) {                int tmp = INF;                if(!(j & 1)) tmp = p[j / 2][i];                ver[i].dp[0][j] = tmp;            }            ver[i].init(2 * n + 1);        }        //columns        for(int i = 1; i <= m; ++i) manacher(b[i], p[i], n);        for(int i = 1; i <= 2 * n + 1; ++i) {            for(int j = 1; j <= 2 * m + 1; ++j) {                int tmp = INF;                if(!(j & 1)) tmp = p[j / 2][i];                hor[i].dp[0][j] = tmp;            }            hor[i].init(2 * m + 1);        }        int ans = 0;        {            int l = 0, r = 150;            while(l <= r) {                int m = l + r >> 1;                if(check(2 * m)) l = m + 1;                else r = m - 1;            }            --l;            ans = max(ans, 2 * l);        }        {            int l = 0, r = 150;            while(l <= r) {                int m = l + r >> 1;                if(check(2 * m + 1)) l = m + 1;                else r = m - 1;            }            --l;            ans = max(ans, 2 * l + 1);        }        printf("%d\n", ans);    }    return 0;}