思路来自 http://blog.csdn.net/u014258433/article/details/76223343
我是用滑动窗口实现的。
代码:
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
using namespace std;
const int maxn = 100;
const long long mod = 1000000007;
int a[maxn + 10][maxn + 10];
int pre[maxn * maxn + 10][maxn + 10];
long long dp[maxn + 10][maxn + 10];
int main()
{
for(int i = 1; i <= maxn; ++i)
{
for(int j = 1; j <= maxn; ++j)
{
if(i == 1 && j == 1)
dp[i][j] = 1;
else if(i == 1)
dp[i][j] = dp[i][j - 1] + j;
else if(j == 1)
dp[i][j] = dp[i - 1][j] + i;
else
dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + i * j;
}
}
int T;
scanf("%d", &T);
while(T--)
{
int n, m;
scanf("%d%d", &n, &m);
memset(pre, 0, sizeof(pre));
for(int i = 1; i <= n; ++i)
{
for(int j = 1; j <= m; ++j)
{
scanf("%d", &a[i][j]);
}
}
long long ans = 0;
for(int i = 1; i <= n; ++i)
{
for(int j = 1; j <= m; ++j)
{
int color = a[i][j];
int l = j, r = j;
int MAX = pre[color][j];
ans += (long long)(i - MAX) * (n - i + 1);
while(l - 1 > 0 && pre[color][l - 1] != i || r + 1 <= m)
{
if(l - 1 > 0 && pre[color][l - 1] != i && r + 1 <= m)
{
int tmp1 = max(pre[color][l - 1], MAX), tmp2 = max(pre[color][r + 1], MAX);
if(tmp1 < tmp2)
{
ans += (long long)(i - tmp1) * (j - l + 2) * (n - i + 1) * (r - j + 1) - (long long)(i - tmp1) * (j - l + 1) * (n - i + 1) * (r - j + 1);
--l;
MAX = tmp1;
}
else
{
ans += (long long)(i - tmp2) * (j - l + 1) * (n - i + 1) * (r - j + 2) - (long long)(i - tmp2) * (j - l + 1) * (n - i + 1) * (r - j + 1);
++r;
MAX = tmp2;
}
}
else if(l - 1 > 0 && pre[color][l - 1] != i)
{
int tmp = max(pre[color][l - 1], MAX);
ans += (long long)(i - tmp) * (j - l + 2) * (n - i + 1) * (r - j + 1) - (long long)(i - tmp) * (j - l + 1) * (n - i + 1) * (r - j + 1);
--l;
MAX = tmp;
}
else
{
int tmp = max(pre[color][r + 1], MAX);
ans += (long long)(i - tmp) * (j - l + 1) * (n - i + 1) * (r - j + 2) - (long long)(i - tmp) * (j - l + 1) * (n - i + 1) * (r - j + 1);
++r;
MAX = tmp;
}
// cout << "l == " << l << " r == " << r << endl;
// cout << "cur_ans == " << ans << endl;
}
// cout << "l == " << l << " r == " << r << endl;
// cout << "ans == " << ans << endl;
pre[color][j] = i;
}
}
printf("%.9f\n", ans * 1.0 / dp[n][m]);
}
return 0;
}
原文:http://www.cnblogs.com/Rhantolk/p/7251671.html