LeetCode 1595 Minimum Cost to Connect Two Groups of Points (動態規劃)
阿新 • • 發佈:2020-10-22
題解: 動態規劃,用二進位制壓縮狀態,注意分析幾種情況,就能推出來正確的狀態轉移方程。
class Solution { public: int dp[12][4096]; int connectTwoGroups(vector<vector<int>>& cost) { int n = cost.size(); if (n == 0) return 0; int m = cost[0].size(); for (int i = 0; i < n; i++) { for (int j = 0; j < (1 << m); j++) { dp[i][j] = INT32_MAX; for (int k = 0; k < m; k++) { if (j & (1 << k)) { if (i != 0 && dp[i - 1][j ^ (1 << k)] != INT32_MAX ) { dp[i][j] = min(dp[i][j], dp[i - 1][j ^ (1 << k)] + cost[i][k]); } if (i != 0 && (dp[i - 1][j] != INT32_MAX)) { dp[i][j] = min(dp[i][j], dp[i-1][j]+cost[i][k]); } if (i == 0 && (j ^ (1 << k)) == 0) { dp[i][j] = cost[i][k]; } else { if (dp[i][j ^ (1 << k)] != INT32_MAX) { dp[i][j] = min(dp[i][j], dp[i][j ^ (1 << k)] + cost[i][k]); } } } } } } return dp[n - 1][(1 << m) - 1]; } };