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題目:uva116 - Unidirectional TSP(記憶化搜尋)
題目大意:給出一個數組,然後可以從第一列任意一行(i, 0)開始走,只能走三個位置(i + 1, 1) (i, 1), (i - 1, 0) 並且這裡預設第一行和最後一行是相連著的,就是當i+ 1或著i - 1超出邊界那麼就到另一頭的邊界。最後輸出字典序最小的路徑。
解題思路:記憶化搜尋。dp【x】【y】 =Min( dp【x + dir【i】【0】】【y + dir【i】【0】】 + mat【x】【y】)。
代碼:
#include <cstdio>#include <cstring>const int N = 15;const int M = 105;const int INF = 0x3f3f3f3f;const int dir[3][2] = {{0, 1}, {1, 1}, {-1, 1}};int mat[N][M];int f[N][M];int path[N][M];int n, m;int Min (const int a, const int b) { return a < b ? a: b; }void init () {for (int i = 0; i <= n; i++)for (int j = 0; j <= m; j++) {f[i][j] = INF;path[i][j] = n;}}int dp (int x, int y) {int& ans = f[x][y];if (y == m)return ans = 0;if (ans != INF)return ans;int nx, ny;int temp;for (int i = 0; i < 3; i++) {nx = x + dir[i][0];ny = y + dir[i][1];if (nx == -1)nx = n - 1;if (nx == n)nx = 0;temp = dp(nx, ny) + mat[x][y]; if (temp <= ans) {if (temp == ans)path[x][y] = Min (path[x][y], nx);elsepath[x][y] = nx;ans = temp;}}return ans;}void printf_ans (int x, int y) {if (y == m - 1)return;printf (" %d", path[x][y] + 1);printf_ans(path[x][y], y + 1);}int main () {while (scanf ("%d%d", &n, &m) != EOF) {for (int i = 0; i < n; i++)for (int j = 0; j < m; j++)scanf ("%d", &mat[i][j]);init();int ans = INF;int temp, r;for (int i = n - 1; i >= 0; i--) {temp = dp(i, 0);if (temp <= ans) {ans = temp;r = i;}}printf ("%d", r + 1);printf_ans(r, 0);printf ("\n%d\n", ans);}return 0;}
uva116 - Unidirectional TSP(記憶化搜尋)