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http://www.lydsy.com/JudgeOnline/problem.php?id=1177
前缀和优化,时间复杂度$O(nm)$
因为数据不全,快速读入会导致RE,切记!
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 1503; int in() { int k = 0, fh = 1; char c = getchar(); for(; c < ‘0‘ || c > ‘9‘; c = getchar()) if (c == ‘-‘) fh = -1; for(; c >= ‘0‘ && c <= ‘9‘; c = getchar()) k = (k << 3) + (k << 1) + c - ‘0‘; return k * fh; } int a[N][N], a1[N][N], a2[N][N], a3[N][N], a4[N][N], s[N][N]; int n, m, k, m1[N][N], m2[N][N], m3[N][N], m4[N][N], h[N], l[N]; void init1() { int S; for(int i = k; i <= n; ++i) for(int j = k; j <= m; ++j) { S = s[i][j] - s[i - k][j] - s[i][j - k] + s[i - k][j - k]; a1[i - k + 1][j - k + 1] = S; a2[i - k + 1][j] = S; a3[i][j - k + 1] = S; a4[i][j] = S; } } void init2() { for(int i = n - k + 1; i >= 1; --i) for(int j = m - k + 1; j >= 1; --j) m1[i][j] = max(a1[i][j], max(m1[i + 1][j], m1[i][j + 1])); for(int i = n - k + 1; i >= 1; --i) for(int j = k; j <= m; ++j) m2[i][j] = max(a2[i][j], max(m2[i + 1][j], m2[i][j - 1])); for(int i = k; i <= n; ++i) for(int j = m - k + 1; j >= 1; --j) m3[i][j] = max(a3[i][j], max(m3[i - 1][j], m3[i][j + 1])); for(int i = k; i <= n; ++i) for(int j = k; j <= m; ++j) m4[i][j] = max(a4[i][j], max(m4[i - 1][j], m4[i][j - 1])); } int main() { scanf("%d%d%d", &n, &m, &k); for(int i = 1; i <= n; ++i) for(int j = 1; j <= m; ++j) { scanf("%d", &a[i][j]); s[i][j] = s[i - 1][j] + s[i][j - 1] + a[i][j] - s[i - 1][j - 1]; } init1(); init2(); int ans = 0, f1, f2, f3, f4; for(int i = k; i <= n - k; ++i) for(int j = k; j <= m - k; ++j) { f1 = m4[i][j] + m3[i][j + 1] + m1[i + 1][1]; f2 = m3[i][j + 1] + m1[i + 1][j + 1] + m2[1][j]; f3 = m2[i + 1][j] + m1[i + 1][j + 1] + m3[i][1]; f4 = m4[i][j] + m2[i + 1][j] + m4[1][j + 1]; ans = max(ans, f1); ans = max(ans, f2); ans = max(ans, f3); ans = max(ans, f4); } for(int i = k + 1; i <= n - (k << 1) + 1; ++i) for(int j = 1; j <= m - k + 1; ++j) h[i] = max(h[i], a1[i][j]); for(int i = k; i <= n - (k << 1); ++i) ans = max(ans, m3[i][1] + m1[i + k + 1][1] + h[i + 1]); for(int j = k + 1; j <= m - (k << 1) + 1; ++j) for(int i = 1; i <= n - k + 1; ++i) l[j] = max(l[j], a1[i][j]); for(int j = k; j <= m - (k << 1); ++j) ans = max(ans, m2[1][j] + m1[1][j + k + 1] + l[j + 1]); printf("%d\n", ans); return 0; }
APIO的题目~
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原文地址:http://www.cnblogs.com/abclzr/p/5693503.html