• LC 1723. Find Minimum Time to Finish All Jobs (dp+二分)


    link

    普通dp:

    class Solution {
    public:
        int n;
        int minimumTimeRequired(vector<int>& jobs, int k) {
            n=jobs.size();
            vector<int> sum(1<<n);
            for(int i=1;i<(1<<n);i++){
                for(int j=0;j<n;j++){
                    if(i&(1<<j)){
                        sum[i]=sum[i-(1<<j)]+jobs[j];
                        break;
                    }
                }
            }
            vector<vector<int>> dp(k,vector<int>(1<<n, INT_MAX));
            for(int i=0;i<(1<<n);i++) dp[0][i]=sum[i];
    
            for(int i=1;i<k;i++){
                for(int j=0;j<(1<<n);j++){
                    if(j==0){
                        dp[i][j]=0;
                        continue;
                    }
                    for(int k=j;k>0;k=((k-1)&j)){
                        dp[i][j]=min(dp[i][j],max(sum[k],dp[i-1][j-k]));
                    }
                }
            }
            return dp[k-1][(1<<n)-1];
        }
    
        
    };
    

    二分:

    class Solution {
    public:
        int n;
        int minimumTimeRequired(vector<int>& jobs, int k) {
            n=jobs.size();
            vector<int> sum(1<<n);
            for(int i=1;i<(1<<n);i++){
                for(int j=0;j<n;j++){
                    if(i&(1<<j)){
                        sum[i]=sum[i-(1<<j)]+jobs[j];
                        break;
                    }
                }
            }
            int left=0;
            int right=0;
            for(int i:jobs){
                left=max(left,i);
                right+=i;
            }
            while(left<=right){
                int mid=left+(right-left)/2;
                int workers=helper(mid,sum);
                if(workers<=k) right=mid-1;
                else left=mid+1;
            }
            return left;
        }
    
        int helper(int limit, vector<int>& sum){
            vector<int> dp(1<<n,INT_MAX);
            dp[0]=0;
            for(int i=1;i<(1<<n);i++){
                for(int j=i;j>0;j= ((j-1)&i)){
                    if(sum[j]<=limit) dp[i]=min(dp[i],1+dp[i-j]);
                }
            }
            return dp[(1<<n)-1];
        }
    };
    
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  • 原文地址:https://www.cnblogs.com/FEIIEF/p/14263704.html
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