HDU 6049-SDJPX is Happy | Multi-university Training Contest 2

Thought comes from FXXL--

A strange place is the third step can not do, that is, ans at least 1, heard that there are questions in the field, and then admin said can not do-(WA I upset)

/*hdu 6049-SDJPX is Happy [enumeration, pruning] | Multi-university Training Contest 2 Test instructions: n-length arrangement n <= 3000 Sort by three steps: 1. The original array is divided into disjoint K-segment 2. Each segment is sorted independently of 3. Select the two segment of the swap ask the number of steps can be successfully sorted by K can be taken to the maximum amount of analysis: pre-processing any segment of the minimum and maximum value of any [L,r] segment can be divided into the maximum number Using f[i,j] to indicate that the so-called effective segment first satisfies max[i,j]-min[i,j] = j-i again satisfies each paragraph in accordance with this increment, that is, the maximum value of the previous paragraph = = after the minimum of a period of [I, J] after the first paragraph for [k, T] enumeration i,j, t = max[i] [j], re-enumeration K, update the answer although the enumeration is very complex, but can be pruned such as requirements: F[i,j] > 0 && i if not 1 min[1,i-1] = 1, max[1, i-1] = i-1 similar to K, t pruning */#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 | Multi-university Training Contest 2