Luogu1169BZOJ1057[ZJOI2007]棋盘制作

一、悬线法

原作者博客 https://rpdreamer.blog.luogu.org/p1169

#include <bits/stdc++.h>
using namespace std;
namespace fdata
{
inline char nextchar()
{
    static const int BS = 1 << 21;
    static char buf[BS], *st, *ed;
    if (st == ed)
        ed = buf + fread(st = buf, 1, BS, stdin);
    return st == ed ? -1 : *st++;
}
inline int poread()
{
    int ret = 0;
    char ch;
    while (!isdigit(ch = nextchar()))
        ;

    do
        ret = ret * 10 + ch - '0';
    while (isdigit(ch = nextchar()));
    return ret;
}
} // namespace fdata
using fdata::poread;
const int MAXN = 2005;
int res[MAXN][MAXN], left_[MAXN][MAXN], right_[MAXN][MAXN], up_[MAXN][MAXN];
int n, m, ans1, ans2;
int main()
{
#ifdef lky233
    system("color FD");
    freopen("testdata.in", "r", stdin);
    freopen("testdata.out", "w", stdout);
#endif
    n = poread();
    m = poread();
    for (register int i = 1; i <= n; ++i)
    {
        for (register int j = 1; j <= m; ++j)
        {
            res[i][j] = poread();
            left_[i][j] = right_[i][j] = j;
            up_[i][j] = 1;
        }
    }
    for (register int i = 1; i <= n; ++i)
        for (register int j = 2; j <= m; ++j)
        {
            if (res[i][j] != res[i][j - 1])
                left_[i][j] = left_[i][j - 1];
        }
    for (register int i = 1; i <= n; ++i)
        for (register int j = m - 1; j > 0; --j)
            if (res[i][j] != res[i][j + 1])
                right_[i][j] = right_[i][j + 1];
    for (register int i = 1; i <= n; ++i)
        for (register int j = 1; j <= m; ++j)
        {
            if (i > 1 && res[i][j] != res[i - 1][j])
            {
                left_[i][j] = max(left_[i][j], left_[i - 1][j]);
                right_[i][j] = min(right_[i][j], right_[i - 1][j]);
                up_[i][j] = up_[i - 1][j] + 1;
            }
            int a = right_[i][j] - left_[i][j] + 1;
            int b = min(a, up_[i][j]);
            ans1 = max(ans1, b * b);
            ans2 = max(ans2, a * up_[i][j]);
        }
    printf("%d\n%d", ans1, ans2);
}

思路：预处理左右边界然后计算长度。

二、单调栈

#include <bits/stdc++.h>
using namespace std;
namespace fdata
{
inline char nextchar()
{
    static const int BS = 1 << 21;
    static char buf[BS], *st, *ed;
    if (st == ed)
        ed = buf + fread(st = buf, 1, BS, stdin);
    return st == ed ? -1 : *st++;
}
inline int poread()
{
    int ret = 0;
    char ch;
    while (!isdigit(ch = nextchar()))
        ;

    do
        ret = ret * 10 + ch - '0';
    while (isdigit(ch = nextchar()));
    return ret;
}
} // namespace fdata
using fdata::poread;
const int MAXN = 2005;
int n, m, top, cur, ans1, ans2;
int stack[MAXN], map[MAXN][MAXN], h[MAXN];
int main()
{
#ifdef lky233
    system("color FD");
    freopen("testdata.in", "r", stdin);
    freopen("testdata.out", "w", stdout);
#endif
    n = poread();
    m = poread();
    for (register int i = 1; i <= n; ++i)
        for (register int j = 1; j <= m; ++j)
            ::map[i][j] = poread();
    for (register int j = 1; j <= n; ++j)
    {
        for (register int i = 1; i <= m; ++i)
            if (j > 1 && ::map[j][i] != ::map[j - 1][i])
                ++h[i];
            else
                h[i] = 1;
        cur = 1;
        while (cur <= m)
        {
            ::stack[0] = cur - 1;
            ::stack[top = 1] = cur++;
            while (cur <= m && ::map[j][cur] != ::map[j][cur - 1])
            {

                while (top && h[::stack[top]] > h[cur])
                {
                    int x = h[::stack[top]];
                    int y = cur - ::stack[top - 1] - 1;
                    int z = min(x, y);
                    ans1 = max(ans1, z * z), ans2 = max(ans2, x * y), --top;
                }
                ::stack[++top] = cur++;
            }
            while (top)
            {
                int x = h[::stack[top]];
                int y = cur - ::stack[top - 1] - 1;
                int z = min(x, y);
                ans1 = max(ans1, z * z);
                ans2 = max(ans2, x * y);
                --top;
            }
        }
    }
    printf("%d\n%d\n", ans1, ans2);
    return 0;
}

 

转载于:https://www.cnblogs.com/Shiina-Rikka/p/11263334.html