UVa 10285 - Longest Run on a Snowboard
2013-05-03 17:54
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记忆化搜索,对d[i][j] 需要考虑 d[i - 1][j], d[i + 1][j], d[i][j - 1], d[i][j + 1]四种情况,但是需要注意边界的控制。
#include <iostream> #include <cstdio> #include <string> #include <cstring> #include <algorithm> using namespace std; int Case, R, C, A[105][105], opt[105][105], d[105][105]; string s; int dp ( int r, int c ) { if ( opt[r][c] ) return d[r][c]; opt[r][c] = 1; int ans1, ans2, ans3, ans4; ans1 = ans2 = ans3 = ans4 = 1; if ( 1 <= c - 1 && c - 1 <= C && A[r][c] > A[r][c - 1] ) ans1 = max ( ans1, dp ( r, c - 1 ) + 1 ); if ( 1 <= c + 1 && c + 1 <= C && A[r][c] > A[r][c + 1] ) ans2 = max ( ans2, dp ( r, c + 1 ) + 1 ); if ( 1 <= r - 1 && r - 1 <= R && A[r][c] > A[r - 1][c] ) ans3 = max ( ans3, dp ( r - 1, c ) + 1 ); if ( 1 <= r + 1 && r + 1 <= R && A[r][c] > A[r + 1][c] ) ans4 = max ( ans4, dp ( r + 1, c ) + 1 ); d[r][c] = max ( ans1, max ( ans2, max ( ans3, ans4 ) ) ); return d[r][c]; } int main ( ) { //freopen ( "input.txt", "r", stdin ); //freopen ( "output.txt", "w", stdout ); cin >> Case; while ( Case-- ) { memset ( A, 0, sizeof ( A ) ); cin >> s >> R >> C; for ( int i = 1; i <= R; ++i ) for ( int j = 1; j <= C; ++j ) cin >> A[i][j]; memset ( opt, 0, sizeof ( opt ) ); memset ( d, 0, sizeof ( d ) ); int Max = -1; for ( int i = 1; i <= R; ++i ) { for ( int j = 1; j <= C; ++j ) { int tmp = dp ( i, j ); //cout << tmp << " "; if ( Max < tmp ) Max = tmp; } //cout << endl; } cout << s << ": " << Max << endl; } }
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