hdu 4826(dp + 记忆化搜索)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4826
思路:dp[x][y][d]表示从方向到达点(x,y)所能得到的最大值,然后就是记忆化了。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define REP(i, a, b) for (int i = (a); i < (b); ++i) #define FOR(i, a, b) for (int i = (a); i <= (b); ++i) using namespace std; const int MAX_N = (100 + 10); const int inf = 1 << 30; int N, M, num[MAX_N][MAX_N]; int dp[MAX_N][MAX_N][3]; void Init() { memset(dp, -1, sizeof(dp)); } int dir[3][2] = {{1, 0}, {-1, 0}, {0, 1}}; int dfs(int x, int y, int d) { if (x == 1 && y == M) return num[x][y]; if (~dp[x][y][d]) return dp[x][y][d]; int res = -inf; REP(i, 0, 3) { int xx = x + dir[i][0], yy = y + dir[i][1]; if ((i == 0 && d == 1) || (i == 1 && d == 0)) continue; if (xx >= 1 && xx <= N && yy >= 1 && yy <= M) { res = max(res, num[x][y] + dfs(xx, yy, i)); } } return dp[x][y][d] = res; } int main() { int Cas, t = 1; scanf("%d", &Cas); while (Cas--) { scanf("%d %d", &N, &M); FOR(i, 1, N) FOR(j, 1, M) scanf("%d", &num[i][j]); Init(); printf("Case #%d:\n%d\n", t++, dfs(1, 1, 0)); } return 0; }