HDU 6049 - Sdjpx Is Happy | 2017 Multi-University Training Contest 2

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思路來源於 FXXL - -

一個比較奇怪的地方就是第三步可以不做，也就是ans至少為1，聽說場內有提問的，然後 admin 說可以不做- - (wa的我心煩)

/*HDU 6049 - Sdjpx Is Happy [ 枚舉,剪枝 ]  |  2017 Multi-University Training Contest 2題意：長度為N的排列 N <= 3000排序分三個步驟：1.原數組分為不相交的K段2.每段都獨立排序3.選擇其中兩段swap問按步驟能成功排序的K能取到的最大是多少分析：先預先處理出任意段的最小值和最大值再處理出任意[l,r]段最多能分成多少段有效段，用f[i,j]表示所謂的有效段首先滿足 Max[i,j]-Min[i,j] = j-i再滿足其中每段依此遞增，即前一段的最大值 == 後一段的最小值-1設需要交換的前一段為[i, j] 後一段為[k, t]枚舉i,j，則 t = Max[i][j]，再枚舉k，更新答案雖然枚舉複雜度大，但可以剪枝比如要求：f[i,j] > 0 && i 如果不為 1 則 Min[1,i-1] = 1, Max[1, i-1] = i-1類似的對k, t剪枝*/#include <bits/stdc++.h>using namespace std;const int N = 3005;int Max[N][N], Min[N][N];int f[N][N];int t, n;int a[N], last[N];void init(){    memset(f, 0, sizeof(f));    int i, j, k;    for (i = 1; i <= n; i++) Max[i][i] = Min[i][i] = a[i];    for (k = 2; k <= n; k++)        for (i = 1; i+k-1 <= n; i++)        {            j = i+k-1;            Max[i][j] = max(Max[i+1][j], Max[i][i]);            Min[i][j] = min(Min[i+1][j], Min[i][i]);        }    for (i = 1; i <= n; i++) f[i][i] = 1, last[i] = i;    for (k = 2; k <= n; k++)        for (i = 1; i+k-1 <= n; i++)        {            j = i+k-1;            if (Max[i][j] - Min[i][j] != j-i) continue;            if (Min[i][last[i]] != Min[i][j]) f[i][j] = 1;            else f[i][j] = f[i][last[i]] + 1;            last[i] = j;        }}int ans;void solve(){    ans = f[1][n];    for (int i = 1; i <= n; i++)    {        if ( i != 1 && (!f[1][i-1] || Max[1][i-1] != i-1)) continue;        for (int j = i; j <= n; j++)        {            if (!f[i][j]) continue;            int t = Max[i][j];            if (t != n && (!f[t+1][n] || Min[t+1][n] != t+1 || Max[t+1][n] != n)) continue;            for (int k = t; k > j; k--)            {                if (!f[k][t] || Min[k][t] != i ) continue;                if (k > j+1)                {                   if (!f[j+1][k-1] || Max[k][t] != Min[j+1][k-1]-1 || Min[i][j] != Max[j+1][k-1]+1) continue;                }                else                {                    if (Max[k][t] != Min[i][j]-1) continue;                }                ans = max(ans, f[1][i-1] + 2 + f[j+1][k-1] + f[t+1][n]);            }        }    }}int main(){    scanf("%d", &t);    while (t--)    {        scanf("%d", &n);        for (int i = 1; i <= n; i++) scanf("%d", a+i);        init();        solve();        printf("%d\n", ans);    }}

　　

HDU 6049 - Sdjpx Is Happy | 2017 Multi-University Training Contest 2