洛谷P1169 [ZJOI2007]棋盘制作 悬线法 动态规划
P1169 [ZJOI2007]棋盘制作
(逼着自己做DP
题意:
给定一个包含0,1的矩阵,求出一个面积最大的正方形矩阵和长方形矩阵,要求矩阵中相邻两个的值不同。
思路:
悬线法。
用途:
解决给定矩阵中满足条件的最大子矩阵
做法:
用一条线(横竖貌似都行)左右移动直到不满足约束条件或者到达边界
定义几个东西:
$left[i][j]$ :代表从(i,j)(i,j)能到达的最左位置
$right[i][j]$ :代表从(i,j)(i,j)能到达的最右位置
$up[i][j]$:代表从(i,j)(i,j)向上扩展最长长度.
递推公式:
$$left[i][j]=max(left[i][j],left[i-1][j])$$
$$right[i][j]=min(right[i][j],right[i-1][j])$$
至于为什么递推公式中考虑上一层的情况?
是因为up数组的定义,up数组代表向上扩展最长长度, 所以需要考虑上一层的情况.
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl '\n' #define OKC ios::sync_with_stdio(false);cin.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) #define max3(a,b,c) max(max(a,b), c); #define min3(a,b,c) min(min(a,b), c); //priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 1e9+7; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黄金分割点 const double tPHI=0.38196601; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } /*-----------------------showtime----------------------*/ const int maxn = 2009; int mp[maxn][maxn]; int lef[maxn][maxn],righ[maxn][maxn],up[maxn][maxn]; int main(){ int n,m; scanf("%d%d", &n, &m); for(int i=1; i<=n; i++){ for(int j=1; j<=m; j++) { scanf("%d", &mp[i][j]); } } for(int i=1; i<=n; i++){ for(int j=1; j<=m; j++){ lef[i][j] = j; if(j>1 && mp[i][j] != mp[i][j-1]) lef[i][j] = lef[i][j-1]; up[i][j] = 1; } } for(int i=1; i<=n; i++){ for(int j=m; j>=1; j--){ righ[i][j] = j; if(j < m && mp[i][j] != mp[i][j+1]) righ[i][j] = righ[i][j+1]; } } int ans1 = 1, ans2 = 1; for(int i=2; i<=n; i++){ for(int j=1; j<=m; j++){ if(mp[i][j] != mp[i-1][j]){ lef[i][j] = max(lef[i][j] , lef[i-1][j]); righ[i][j] = min(righ[i][j], righ[i-1][j]); up[i][j] = up[i-1][j] + 1; } int a = righ[i][j] - lef[i][j] + 1; int b = up[i][j]; // cout<<a<<" "<<b<<endl; ans2 = max(a * b, ans2); int t = min(a, b); ans1 = max(t * t, ans1); } } printf("%d\n%d\n", ans1, ans2); return 0; }
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