Experiment 3: Greedy Algorithm: Traveling Salesman Problem (TSP)
Problem description: There are several cities, and the transportation fee between any two cities is determined. Now a traveling salesman starts from a certain city and passes through each city only once. Finally, Go back to the original city and determine a route to minimize travel expenses.
Experiment 4: Eight Queens Problem.
Problem description: N2 squares are arranged in a square with N rows and N columns, which is called an N-yuan chessboard. Place N queens on the N-yuan chessboard. If two queens are located in the same row or column or the same diagonal on the N-yuan chessboard On the line (slope ±1), it is said that they are attacking each other. Try to design an algorithm to find all the layouts that make the N queens not attack each other.
Problem-solving ideas;
-
First of all: write a function to determine whether the position of the k-th queen meets the requirements.
Because the loop increment variable i can be restricted to not be in the same row, in this case, X[i] is used to restrict the not in the same column, that is: x≠X[i] is
on the same main diagonal: the row-column values are all equal iX[i ]=kX[k] is
on the same anti-diagonal line: the values of row + column are all equal i+X[i]=k+X[k]
can be derived from the general formula of main diagonal and anti-diagonal line: |X[ i]-X[k]|=|ik|, if it is not on a diagonal, you need to ensure that ABS (|X[i]-X[k]|)≠ABS(ik)
from i=1 to k if If the queen is not in the same row or column or on the diagonal, it is true, otherwise it is not true. -
Next, I simulated the execution of the main function according to the pseudo code in the algorithm book.
-
The main function idea is as follows:
step1: define row variable k and column variable X[k], k assign initial value 1, X[k] assign initial value 0
step2: define while (k>0) loop, because k initial value is 1 , So when the backtracking comes back to k=0, the program ends.
Step3: Increment X[k] by 1, and define a while (X[k]<=n also substandard) loop to determine whether the position of the queen meets the standard. At this time, the column will be incremented by 1.
step4: If the position is found and X[k]<=n, first judge whether the n queens have positions (k=n), if there are positions, then print the array position, otherwise k++, the column is cleared. If X[k]>n, then k is decremented for backtracking.
The idea is very clear and specific, but there is still a problem in writing the code. The problem is that {} is not added after an else, and another problem is that there is no return to mark the value after the function is called.
code show as below:
#include <stdio.h>
#include <math.h>
int place(int k);
int k=1,x[100]={
0};
int main(){
int n,sum=0;
printf("请输入皇后数量：");
scanf("%d",&n);
while(k>0){
x[k]++;
while(x[k]<=n&&place(k)==0){
x[k]++;
}
if(x[k]<=n){
if(n==k){
//printf("成立\n");
sum++;
}
else{
k++;x[k]=0;
}
}
else k--;
}
printf("sum=%d\n",sum);
}
int place(int k){
int i=1,t=1;
while(i<k){
if(abs(x[i]-x[k])==abs(i-k))
t=0;
if(x[i]==x[k])
t=0;
i++;
}
return t;
}
The code to print the position of the queen is as follows:
#include <stdio.h>
#include <math.h>
int place(int k);
int k=1,x[100]={
0};
int main(){
int n,sum=0;
printf("请输入皇后数量：");
scanf("%d",&n);
while(k>0){
x[k]++;
while(x[k]<=n&&place(k)==0){
x[k]++;
}
if(x[k]<=n){
if(n==k){
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
if(x[i]==j)
printf("#");
else printf("*");
}
printf("\n");
}
printf("\n");
sum++;
}
else{
k++;x[k]=0;
}
}
else k--;
}
printf("sum=%d\n",sum);
}
int place(int k){
int i=1,t=1;
while(i<k){
if(abs(x[i]-x[k])==abs(i-k))
t=0;
if(x[i]==x[k])
t=0;
i++;
}
return t;
}