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N皇后問題:基於區域性搜尋策略的高效演算法

#include<iostream>
#include<string.h>
#include<time.h>
using namespace std;

#define TIMES 10  //測試輪數
#define MAX_SIZE 10000  //最大可求解的皇后數量

int n;  //待輸入的皇后數量
int board[MAX_SIZE];  //記錄棋盤狀況的陣列
int label[MAX_SIZE];  //標記陣列
//記錄衝突的陣列(使用暴力迴圈方法檢測衝突時可以不需要定義這兩個陣列)
int ru[MAX_SIZE * 2];  //右上
int rd[MAX_SIZE * 2];  //右下

//生成x個不重複的隨機數,並放入board陣列
void RandArray(int x) {
    int i, temp;
    memset(label, 0, sizeof(int) * (n + 1));
    for (i = 0; i < n; i++) {
        do {
            temp = rand() % x;
        } while (label[temp] == 1);
        board[i] = temp;
        label[temp] = 1;
    }
}

/*************************************************/
/*int Conflict() {  //使用暴力迴圈方法計算衝突
    int i, j, r = 0;
    for (i = 0; i < n - 1; i++)
        for (j = i + 1; j < n; j++) {
            if (j - i == abs(board[j] - board[i]))
                r++;
        }
    return r;
}*/
int Conflict() {  //使用陣列方法計算衝突
    int i, r = 0;
    memset(ru, 0, sizeof(ru));
    memset(rd, 0, sizeof(rd));
    for (i = 0; i < n; i++) {
        ru[board[i] - i + n]++;
        rd[board[i] + i]++;
    }
    for (i = 0; i < 2 * n; i++) {
        if (ru[i] > 1) r += ru[i] - 1;
        if (rd[i] > 1) r += rd[i] - 1;
    }
    return r;
}
/*************************************************/

int main() {
    srand((unsigned) time(NULL));
    int i, j, temp, now, t1, t2, fnow, sumtime = 0, avet;
    cout << "\n輸入棋盤大小(皇后個數):";
    cin >> n;
    cout << endl;
    for (avet = 0; avet < TIMES; avet++) {  //測試TIMES輪
        t1 = clock();
        while (1) {  //區域性搜尋部分
            RandArray(n);
            now = Conflict();
            next:
            if (now == 0) goto over;
            for (i = 0; i < n - 1; i++) {
                for (j = i + 1; j < n; j++) {  //可以通過改變j的步長來實現部分交換,但經過測試發現,效率都不如單步長方法
                    temp = board[i];
                    board[i] = board[j];
                    board[j] = temp;
                    fnow = Conflict();
                    if (fnow < now) {
                        now = fnow;
                        goto next;
                    }
                    temp = board[i];
                    board[i] = board[j];
                    board[j] = temp;
                }
            }
        }
        over:
        t2 = clock();
        /*for (i = 0; i < n; i++) {  //列印皇后在棋盤中的分佈
            for (j = 0; j < n; j++) {
                if (board[i] == j) cout << " X";
                else cout << " +";
            }
            cout << endl;
        }*/
        sumtime += (t2 - t1);
        cout << "第 " << avet + 1 << " 輪耗費時間:" << (t2 - t1) << " ms\n";
    }
    cout << "\n" << avet << " 輪平均耗費時間為:" << sumtime / avet << " ms\n";
    return 0;
}

列印棋盤:

演算法效率: