leetcode 1723. 完成所有工作的最短时间【DFS+剪枝 | 状压DP】

DFS+剪枝

超时

​​题解​​

题解看了一半 学到了用当前得到的最好的结果进行剪枝，但是还是超时
后来发现当有多个空闲的工人 或者多个工人当前已经总用时相同时重复计算了问题
因为工人是没有区分的

假设 工人用时是这样的[0,0,0,0,0,1,1]
那么在0和1的这些工人里面调一个开分支即可

#define debug(x) cout<<#x<<": "<<(x)<<endl;

class Solution {
public:
    int ret = INT_MAX;
    vector<int>j;
    int k;
    int minimumTimeRequired(vector<int>& j, int k) {
        vector<int> p(k,0);
        this->j = j;
        this->k = k;
        dfs(p,0);
        return ret;
    }

    bool dfs(vector<int>w,int cur){
        int m = 0;
        debug(cur)
        if(cur == j.size()){
            int ma = -1;
            for(int i=0;i<w.size();i++){
                ma = max(ma,w[i]);
            }
            ret = min(ret,ma);
            return true;
        }

        sort(w.begin(),w.end());
        
        for(int i=0;i< w.size();i++){
            if( w[i] + j[cur] < ret){
                w[i]+=j[cur];
                dfs(w,cur+1);
                w[i]-=j[cur];
            }
        }
        return true;
    }
};

加入剪枝
if(i>0 && w[i]==w[i-1]){ continue;}
即可通过
因为将工人进行了排序 所以直接比较i和i-1即可

#define debug(x) cout<<#x<<": "<<(x)<<endl;

class Solution {
public:
    int ret = INT_MAX;
    vector<int>j;
    int k;
    int minimumTimeRequired(vector<int>& j, int k) {
        vector<int> p(k,0);
        this->j = j;
        this->k = k;
        dfs(p,0);
        return ret;
    }

    bool dfs(vector<int>w,int cur){
        
        if(cur == j.size()){
            int ma = -1;
            for(int i=0;i<w.size();i++){
                ma = max(ma,w[i]);
            }
            ret = min(ret,ma);
            return true;
        }

        sort(w.begin(),w.end());
        
        for(int i=0;i< w.size();i++){
            if( w[i] + j[cur] < ret){
                if(i>0 && w[i]==w[i-1]){
                    continue;
                }
                w[i]+=j[cur];
                dfs(w,cur+1);
                w[i]-=j[cur];
            }
        }
        return true;
    }
};

状态压缩DP

#define debug(x) cout<<#x<<": "<<(x)<<endl;

class Solution {
public:

    bool print2(int a){
        vector<bool>b;
        while(a>0){
            b.insert(b.begin(), (a&1) );
            a>>=1;
        }
        while(b.size()<16){
            b.insert(b.begin(), 0);
        }
        for(int i=0;i<16;++i){
            cout<<b[i]<<" ";
        }
        cout<<endl;
        return true;
    }
    int last1(int n){
        if(n==0){
            return -1;
        }
        int i=0;
        while(n>0){
            if(n&1){
                return i;
            }
            i++;
            n>>=1;
        }
        return i;
    }
    int minimumTimeRequired(vector<int>& j, int k) {
        int n = j.size();
        vector<vector<int>>dp(k,vector<int>(1<<n,INT_MAX));
        vector<int>sum(1<<n,0);
        
        for(int i=1;i<(1<<n);++i){
            int la = last1(i);
            sum[i] = j[la] + sum[i- (1<<la)];
        }
        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++){
                for(int t = j ;t>0;t = j &(t-1) ){
                    dp[i][j] = min(dp[i][j], max( dp[i-1][t], sum[ j - t]));
                }
            }
        }
        return dp[k-1][(1<<n)-1];
    }
};

取二进制 1111的子集
1 1 1 1
1 1 1 0
1 1 0 1
1 1 0 0
1 0 1 1
1 0 1 0
1 0 0 1
1 0 0 0
0 1 1 1
0 1 1 0
0 1 0 1
0 1 0 0
0 0 1 1
0 0 1 0
0 0 0 1
0 0 0 0

#include <iostream>
#include <math.h>
#include <vector>

using namespace std;
#define debug(x) cout<< #x <<": "<< (x) <<endl;

int maxBit = 4;
bool print2(int n) {
    vector<int> b;
    while (n>1){
        b.insert(b.begin(),n&1);
        n >>= 1;
    }
    b.insert(b.begin(), n & 1);

    while (b.size()< maxBit){
        b.insert(b.begin(), 0);
    }
    for (int i = 0; i < b.size(); i++) {
        cout<< b[i] << " ";
    }
    cout << endl;
    return true;
}

int main()
{

    int n = (1 << maxBit);

    while (n-- >0){
        print2(n);
    }

    debug("-----")
    int j = 13;
    int i = 13;
    while (i > 0) {
        print2(i);
        i = (i - 1) & j;
        
    }

    return 0;
}

结果

1 1 1 1
1 1 1 0
1 1 0 1
1 1 0 0
1 0 1 1
1 0 1 0
1 0 0 1
1 0 0 0
0 1 1 1
0 1 1 0
0 1 0 1
0 1 0 0
0 0 1 1
0 0 1 0
0 0 0 1
0 0 0 0
"-----": -----
1 1 0 1
1 1 0 0
1 0 0 1
1 0 0 0
0 1 0 1
0 1 0 0
0 0 0 1

要是1不挨着，这个时候不能单纯-1
要和原来的数&一下 如13