Leetcode 1986. Minimum Number of Work Sessions to Finish the Tasks
Woodyiiiiiii opened this issue · comments
Bitmask + up-bottom dfs
讲解：
class Solution {
public int minSessions(int[] tasks, int sessionTime) {
int n = tasks.length;
Integer[][] dp = new Integer[1 << n][sessionTime];
return dfs(tasks, n, dp, (1 << n) - 1, 0, sessionTime);
}
private int dfs(int[] tasks, int n, Integer[][] dp, int mask, int remainTime, final int sessionTime) {
if (mask == 0) {
return 0;
}
if (dp[mask][remainTime] != null) {
return dp[mask][remainTime];
}
int ans = n;
for (int i = 0; i < n; ++i) {
if ((mask & (1 << i)) == 0) {
continue;
}
int newMask = mask ^ (1 << i);
if (tasks[i] <= remainTime) {
ans = Math.min(ans, dfs(tasks, n, dp, newMask, remainTime - tasks[i], sessionTime));
} else {
ans = Math.min(ans, 1 + dfs(tasks, n, dp, newMask, sessionTime - tasks[i], sessionTime));
}
}
return dp[mask][remainTime] = ans;
}
}
class Solution {
public int minSessions(int[] tasks, int sessionTime) {
int n = tasks.length;
Pair[] dp = new Pair[1 << n];
return dfs(tasks, n, dp, (1 << n) - 1, sessionTime).getKey();
}
private Pair dfs(int[] tasks, int n, Pair[] dp, int mask, final int sessionTime) {
if (mask == 0) {
return new Pair(0, 0);
}
if (dp[mask] != null) {
return dp[mask];
}
int ans = n, remain = 0;
for (int i = 0; i < n; ++i) {
if ((mask & (1 << i)) == 0) {
continue;
}
int newMask = mask ^ (1 << i);
Pair p = dfs(tasks, n, dp, newMask, sessionTime);
int sessionCnt = p.getKey(), remainTime = p.getValue();
if (tasks[i] > remainTime) {
sessionCnt++;
remainTime = sessionTime - tasks[i];
} else {
remainTime -= tasks[i];
}
if (sessionCnt < ans || (sessionCnt == ans && remainTime > remain)) {
ans = sessionCnt;
remain = remainTime;
}
}
return dp[mask]= new Pair(ans, remain);
}
static class Pair {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
int getKey() {
return x;
}
int getValue() {
return y;
}
}
}