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2021-10-31 TSP旅行商问题

2021-10-31 TSP旅行商问题

作者: 程博颖 | 来源:发表于2021-10-31 15:30 被阅读0次
    image.png
    import java.util.ArrayList;
    import java.util.List;
    
    public class Code01_TSP {
    
        public static int t1(int[][] matrix) {
            int N = matrix.length; // 0...N-1
            // set
            // set.get(i) != null i这座城市在集合里
            // set.get(i) == null i这座城市不在集合里
            List<Integer> set = new ArrayList<>();
            for (int i = 0; i < N; i++) {
                set.add(1);
            }
            return func1(matrix, set, 0);
        }
    
        // 任何两座城市之间的距离,可以在matrix里面拿到
        // set中表示着哪些城市的集合,
        // start这座城一定在set里,
        // 从start出发,要把set中所有的城市过一遍,最终回到0这座城市,最小距离是多少
        public static int func1(int[][] matrix, List<Integer> set, int start) {
            int cityNum = 0;
            for (int i = 0; i < set.size(); i++) {
                if (set.get(i) != null) {
                    cityNum++;
                }
            }
            if (cityNum == 1) {
                return matrix[start][0];
            }
            // 不只start这一座城
            set.set(start, null);
            int min = Integer.MAX_VALUE;
            for (int i = 0; i < set.size(); i++) {
                if (set.get(i) != null && i != start) {
                    // start -> i i... -> 0
                    int cur = matrix[start][i] + func1(matrix, set, i);
                    min = Math.min(min, cur);
                }
            }
            set.set(start, 1);
            return min;
        }
    
        public static int t2(int[][] matrix) {
            int N = matrix.length; // 0...N-1
            // 7座城 1111111
            int allCity = (1 << N) - 1;
            return f2(matrix, allCity, 0);
        }
    
        // 任何两座城市之间的距离,可以在matrix里面拿到
        // set中表示着哪些城市的集合,
        // start这座城一定在set里,
        // 从start出发,要把set中所有的城市过一遍,最终回到0这座城市,最小距离是多少
        public static int f2(int[][] matrix, int cityStatus, int start) {
            // cityStatus == cityStatux & (~cityStaus + 1)
    
            if (cityStatus == (cityStatus & (~cityStatus + 1))) {
                return matrix[start][0];
            }
    
            // 把start位的1去掉,
            cityStatus &= (~(1 << start));
            int min = Integer.MAX_VALUE;
            // 枚举所有的城市
            for (int move = 0; move < matrix.length; move++) {
                if (move != start && (cityStatus & (1 << move)) != 0) {
                    int cur = matrix[start][move] + f2(matrix, cityStatus, move);
                    min = Math.min(min, cur);
                }
            }
            cityStatus |= (1 << start);
            return min;
        }
    
        public static int t3(int[][] matrix) {
            int N = matrix.length; // 0...N-1
            // 7座城 1111111
            int allCity = (1 << N) - 1;
            int[][] dp = new int[1 << N][N];
            for (int i = 0; i < (1 << N); i++) {
                for (int j = 0; j < N; j++) {
                    dp[i][j] = -1;
                }
            }
            return f3(matrix, allCity, 0, dp);
        }
    
        // 任何两座城市之间的距离,可以在matrix里面拿到
        // set中表示着哪些城市的集合,
        // start这座城一定在set里,
        // 从start出发,要把set中所有的城市过一遍,最终回到0这座城市,最小距离是多少
        public static int f3(int[][] matrix, int cityStatus, int start, int[][] dp) {
            if (dp[cityStatus][start] != -1) {
                return dp[cityStatus][start];
            }
            if (cityStatus == (cityStatus & (~cityStatus + 1))) {
                dp[cityStatus][start] = matrix[start][0];
            } else {
                // 把start位的1去掉,
                cityStatus &= (~(1 << start));
                int min = Integer.MAX_VALUE;
                // 枚举所有的城市
                for (int move = 0; move < matrix.length; move++) {
                    if (move != start && (cityStatus & (1 << move)) != 0) {
                        int cur = matrix[start][move] + f3(matrix, cityStatus, move, dp);
                        min = Math.min(min, cur);
                    }
                }
                cityStatus |= (1 << start);
                dp[cityStatus][start] = min;
            }
            return dp[cityStatus][start];
        }
    
        public static int t4(int[][] matrix) {
            int N = matrix.length; // 0...N-1
            int statusNums = 1 << N;
            int[][] dp = new int[statusNums][N];
    
            for (int status = 0; status < statusNums; status++) {
                for (int start = 0; start < N; start++) {
                    if ((status & (1 << start)) != 0) {
                        if (status == (status & (~status + 1))) {
                            dp[status][start] = matrix[start][0];
                        } else {
                            int min = Integer.MAX_VALUE;
                            // start 城市在status里去掉之后,的状态
                            int preStatus = status & (~(1 << start));
                            // start -> i
                            for (int i = 0; i < N; i++) {
                                if ((preStatus & (1 << i)) != 0) {
                                    int cur = matrix[start][i] + dp[preStatus][i];
                                    min = Math.min(min, cur);
                                }
                            }
                            dp[status][start] = min;
                        }
                    }
                }
            }
            return dp[statusNums - 1][0];
        }
    
        // matrix[i][j] -> i城市到j城市的距离
        public static int tsp1(int[][] matrix, int origin) {
            if (matrix == null || matrix.length < 2 || origin < 0 || origin >= matrix.length) {
                return 0;
            }
            // 要考虑的集合
            ArrayList<Integer> cities = new ArrayList<>();
            // cities[0] != null 表示0城在集合里
            // cities[i] != null 表示i城在集合里
            for (int i = 0; i < matrix.length; i++) {
                cities.add(1);
            }
            // null,1,1,1,1,1,1
            // origin城不参与集合
            cities.set(origin, null);
            return process(matrix, origin, cities, origin);
        }
    
        // matrix 所有距离,存在其中
        // origin 固定参数,唯一的目标
        // cities 要考虑的集合,一定不含有origin
        // 当前来到的城市是谁,cur
        public static int process(int[][] matrix, int aim, ArrayList<Integer> cities, int cur) {
            boolean hasCity = false; // 集团中还是否有城市
            int ans = Integer.MAX_VALUE;
            for (int i = 0; i < cities.size(); i++) {
                if (cities.get(i) != null) {
                    hasCity = true;
                    cities.set(i, null);
                    // matrix[cur][i] + f(i, 集团(去掉i) )
                    ans = Math.min(ans, matrix[cur][i] + process(matrix, aim, cities, i));
                    cities.set(i, 1);
                }
            }
            return hasCity ? ans : matrix[cur][aim];
        }
    
        // cities 里,一定含有cur这座城
        // 解决的是,集合从cur出发,通过集合里所有的城市,最终来到aim,最短距离
        public static int process2(int[][] matrix, int aim, ArrayList<Integer> cities, int cur) {
            if (cities.size() == 1) {
                return matrix[cur][aim];
            }
            cities.set(cur, null);
            int ans = Integer.MAX_VALUE;
            for (int i = 0; i < cities.size(); i++) {
                if (cities.get(i) != null) {
                    int dis = matrix[cur][i] + process2(matrix, aim, cities, i);
                    ans = Math.min(ans, dis);
                }
            }
            cities.set(cur, 1);
            return ans;
        }
    
        public static int tsp2(int[][] matrix, int origin) {
            if (matrix == null || matrix.length < 2 || origin < 0 || origin >= matrix.length) {
                return 0;
            }
            int N = matrix.length - 1; // 除去origin之后是n-1个点
            int S = 1 << N; // 状态数量
            int[][] dp = new int[S][N];
            int icity = 0;
            int kcity = 0;
            for (int i = 0; i < N; i++) {
                icity = i < origin ? i : i + 1;
                // 00000000 i
                dp[0][i] = matrix[icity][origin];
            }
            for (int status = 1; status < S; status++) {
                // 尝试每一种状态 status = 0 0 1 0 0 0 0 0 0
                // 下标 8 7 6 5 4 3 2 1 0
                for (int i = 0; i < N; i++) {
                    // i 枚举的出发城市
                    dp[status][i] = Integer.MAX_VALUE;
                    if ((1 << i & status) != 0) {
                        // 如果i这座城是可以枚举的,i = 6 , i对应的原始城的编号,icity
                        icity = i < origin ? i : i + 1;
                        for (int k = 0; k < N; k++) { // i 这一步连到的点,k
                            if ((1 << k & status) != 0) { // i 这一步可以连到k
                                kcity = k < origin ? k : k + 1; // k对应的原始城的编号,kcity
                                dp[status][i] = Math.min(dp[status][i], dp[status ^ (1 << i)][k] + matrix[icity][kcity]);
                            }
                        }
                    }
                }
            }
            int ans = Integer.MAX_VALUE;
            for (int i = 0; i < N; i++) {
                icity = i < origin ? i : i + 1;
                ans = Math.min(ans, dp[S - 1][i] + matrix[origin][icity]);
            }
            return ans;
        }
    
        public static int[][] generateGraph(int maxSize, int maxValue) {
            int len = (int) (Math.random() * maxSize) + 1;
            int[][] matrix = new int[len][len];
            for (int i = 0; i < len; i++) {
                for (int j = 0; j < len; j++) {
                    matrix[i][j] = (int) (Math.random() * maxValue) + 1;
                }
            }
            for (int i = 0; i < len; i++) {
                matrix[i][i] = 0;
            }
            return matrix;
        }
    
        public static void main(String[] args) {
            int len = 10;
            int value = 100;
            System.out.println("功能测试开始");
            for (int i = 0; i < 20000; i++) {
                int[][] matrix = generateGraph(len, value);
                int origin = (int) (Math.random() * matrix.length);
                int ans1 = t3(matrix);
                int ans2 = t4(matrix);
                int ans3 = tsp2(matrix, origin);
                if (ans1 != ans2 || ans1 != ans3) {
                    System.out.println("fuck");
                }
            }
            System.out.println("功能测试结束");
    
            len = 22;
            System.out.println("性能测试开始,数据规模 : " + len);
            int[][] matrix = new int[len][len];
            for (int i = 0; i < len; i++) {
                for (int j = 0; j < len; j++) {
                    matrix[i][j] = (int) (Math.random() * value) + 1;
                }
            }
            for (int i = 0; i < len; i++) {
                matrix[i][i] = 0;
            }
            long start;
            long end;
            start = System.currentTimeMillis();
            t4(matrix);
            end = System.currentTimeMillis();
            System.out.println("运行时间 : " + (end - start) + " 毫秒");
            System.out.println("性能测试结束");
    
        }
    
    }
    
    

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