您的位置:首页 > 大数据 > 人工智能

矩阵快速幂 CodeForces - 582B Once Again...

2015-10-05 20:34 369 查看
建出转移矩阵...然后做矩阵快速幂就行了....

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

const int maxn = 105;
const int INF = 0x3f3f3f3f;

int mat[maxn][maxn];
int res[maxn][maxn];
int mid[maxn][maxn];
int a[maxn];
int n, m;

void calc(int b)
{
memset(res, 0, sizeof res);
while(b) {
if(b % 2) {
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++) {
mid[i][j] = -INF;
for(int k = 1; k <= n; k++)
mid[i][j] = max(mid[i][j], res[i][k] + mat[k][j]);
}
memcpy(res, mid, sizeof mid);
}
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++) {
mid[i][j] = -INF;
for(int k = 1; k <= n; k++)
mid[i][j] = max(mid[i][j], mat[i][k] + mat[k][j]);
}
memcpy(mat, mid, sizeof mid);
b /= 2;
}
}

void work()
{
scanf("%d%d", &n, &m);

for(int i = 1; i <= n; i++) scanf("%d", &a[i]);

for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
mat[i][j] = 0;

for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++) {
if(a[i] <= a[j]) {
mat[i][j] = 1;
for(int k = 1; k < j; k++)
if(a[k] <= a[j])
mat[i][j] = max(mat[i][j], mat[i][k] + 1);
}
else mat[i][j] = -INF;
}
calc(m);

int ans = 0;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
ans = max(ans, res[i][j]);
printf("%d\n", ans);
}

int main()
{
//freopen("data", "r", stdin);
work();

return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: