使用c语言，解决模拟退火解决n皇后问题，写出代码

在C语言中，使用模拟退火算法求解N皇后问题是常见的优化问题实例。下面是一个简单的示例代码，它实现了基本的模拟退火过程来寻找N皇后问题的一个解决方案：

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

// 检查在给定列和对角线位置上是否可以放置皇后
int isSafe(int row, int col, int N, int board[][N]) {
    // 对角线上检查
    for (int i = 0; i < row; ++i)
        if (board[i][col] == 1 || abs(i - row) == abs(col - board[i][col]))
            return 0;
    return 1;
}

// 初始化棋盘状态
void initBoard(int N, int board[][N]) {
    for (int i = 0; i < N; ++i)
        for (int j = 0; j < N; ++j)
            board[i][j] = 0;
}

// 交换两个皇后的位置
void swap(int row1, int col1, int row2, int col2, int board[][N]) {
    int temp = board[row1][col1];
    board[row1][col1] = board[row2][col2];
    board[row2][col2] = temp;
}

// 扩展函数：模拟退火步骤
void simulatedAnnealing(int N, double initialTemp, double coolingRate, int maxIterations, int board[][N]) {
    srand(time(0));
    int currentEnergy = -999999; // 目前的状态能量
    int bestRow[N], bestCol[N]; // 最佳皇后位置

    while (initialTemp > 1e-6 && maxIterations--) {
        int i = rand() % N;
        int j = rand() % N;

        if (isSafe(i, j, N, board)) {
            swap(i, bestCol[i], j, bestCol[j], board);
            int newEnergy = calculateEnergy(N, board); // 计算新状态的能量
            double acceptanceProb = exp((currentEnergy - newEnergy) / initialTemp);

            if (acceptanceProb > random()) { // 接受概率大于随机数，则接受新状态
                currentEnergy = newEnergy;
                for (int k = 0; k < N; ++k) {
                    bestRow[k] = i;
                    bestCol[k] = j;
                }
            } else { // 否则，尝试恢复旧状态
                swap(i, j, bestRow[i], bestCol[i], board);
            }
        }

        initialTemp *= coolingRate;
    }

    printf("\nSolution found with the following positions:\n");
    for (int i = 0; i < N; ++i) {
        printf("Q at (%d,%d)\n", i, bestCol[i]);
    }
}

// 能量计算函数，这里假设每行、每列和每个对角线只有一个皇后时能量最小，计为-1，否则为0
int calculateEnergy(int N, int board[][N]) {
    int energy = 0;
    for (int i = 0; i < N; ++i)
        for (int j = 0; j < N; ++j)
            energy += (board[i][j] == 1 ? 1 : 0);
    return energy;
}

int main() {
    int N;
    printf("Enter the size of the chessboard (N): ");
    scanf("%d", &N);
    
    int board[N][N];

    simulatedAnnealing(N, 100, 0.95, 1000, board);

    return 0;
}