需要Tkinter代码用于python GUI，以实现分支绑定和动态编程方法中的TSP。。

Question

需要Tkinter代码用于python GUI，以实现分支绑定和动态编程方法中的TSP。。 下面的代码运行良好，我需要一个图形用户界面平台为我的项目演示明天。 我试过用tkinter编程，但没能很好地处理代码。 在GUI中，我需要文本框来读取no:of cities和n*n矩阵。输出屏幕显示路径成本和最佳路径。你们能帮我吗？提前谢谢。在

import itertools

    INF = 100000000

    best_cost = 0

def reduce(size, w, row, col, rowred, colred):
rvalue = 0
for i in range(size):
    temp = INF
    for j in range(size):
        temp = min(temp, w[row[i]][col[j]])
    if temp > 0:
        for j in range(size):
            if w[row[i]][col[j]] < INF:
                w[row[i]][col[j]] -= temp
        rvalue += temp
    rowred[i] = temp
for j in range(size):
    temp = INF
    for i in range(size):
        temp = min(temp, w[row[i]][col[j]])
    if temp > 0:
        for i in range(size):
            if w[row[i]][col[j]] < INF:
                w[row[i]][col[j]] -= temp
        rvalue += temp
    colred[j] = temp
return rvalue


def bestEdge(size, w, row, col):
mosti = -INF
ri = 0
ci = 0
for i in range(size):
    for j in range(size):
        if not w[row[i]][col[j]]:
            minrowwelt = INF
            zeroes = 0
            for k in range(size):
                if not w[row[i]][col[k]]:
                    zeroes += 1
                else:
                    minrowwelt = min(minrowwelt, w[row[i]][col[k]])
            if zeroes > 1: minrowwelt = 0
            mincolwelt = INF
            zeroes = 0
            for k in range(size):
                if not w[row[k]][col[j]]:
                    zeroes += 1
                else:
                    mincolwelt = min(mincolwelt, w[row[k]][col[j]])
            if zeroes > 1: mincolwelt = 0
            if minrowwelt + mincolwelt > mosti:
                mosti = minrowwelt + mincolwelt
                ri = i
                ci = j
return mosti, ri, ci


def explore(n, w, edges, cost, row, col, best, fwdptr, backptr):
global best_cost

colred = [0 for _ in range(n)]
rowred = [0 for _ in range(n)]
size = n - edges
cost += reduce(size, w, row, col, rowred, colred)
if cost < best_cost:
    if edges == n - 2:
        for i in range(n): best[i] = fwdptr[i]
        if w[row[0]][col[0]] >= INF:
            avoid = 0
        else:
            avoid = 1
        best[row[0]] = col[1 - avoid]
        best[row[1]] = col[avoid]
        best_cost = cost
    else:
        mostv, rv, cv = bestEdge(size, w, row, col)
        lowerbound = cost + mostv
        fwdptr[row[rv]] = col[cv]
        backptr[col[cv]] = row[rv]
        last = col[cv]
        while fwdptr[last] != INF: last = fwdptr[last]
        first = row[rv]
        while backptr[first] != INF: first = backptr[first]
        colrowval = w[last][first]
        w[last][first] = INF
        newcol = [INF for _ in range(size)]
        newrow = [INF for _ in range(size)]
        for i in range(rv): newrow[i] = row[i]
        for i in range(rv, size - 1): newrow[i] = row[i + 1]
        for i in range(cv): newcol[i] = col[i]
        for i in range(cv, size - 1): newcol[i] = col[i + 1]
        explore(n, w, edges + 1, cost, newrow, newcol, best, fwdptr, backptr)
        w[last][first] = colrowval
        backptr[col[cv]] = INF
        fwdptr[row[rv]] = INF
        if lowerbound < best_cost:
            w[row[rv]][col[cv]] = INF
            explore(n, w, edges, cost, row, col, best, fwdptr, backptr)
            w[row[rv]][col[cv]] = 0

for i in range(size):
    for j in range(size):
        w[row[i]][col[j]] = w[row[i]][col[j]] + rowred[i] + colred[j]

#  code for branch & bound 
def bb(w,size):
global best_cost
for i in range(size):
    a[i][i]=INF
print('The cost matrix:')
for i in range(size):
    print(a[i])
col = [i for i in range(size)]
row = [i for i in range(size)]
best = [0 for _ in range(size)]
route = [0 for _ in range(size)]
fwdptr = [INF for _ in range(size)]
backptr = [INF for _ in range(size)]
best_cost = INF

explore(size, w, 0, 0, row, col, best, fwdptr, backptr)

index = 0
for i in range(size):
    route[i] = index
    index = best[index]
index = []
cost = 0

for i in range(size):
    if i != size - 1:
        src = route[i]
        dst = route[i + 1]
    else:
        src = route[i]
        dst = 0
    cost += w[src][dst]
    index.append(src+1)
index.append(1)
print('B&B Minimum cost:',cost)
print('B&B Optimal path:',index)
return

#   code for dynamic programming
def dp(A,N):
v=[x for x in range(1,N)]
s=tuple(itertools.permutations(v,N-1))

l=len(s)
C=[]
for i in range(l):
    t=s[i]
    c=A[0][t[0]]
    for j in range(1,len(t)):
        c+=A[t[j-1]][t[j]]
    c+=A[t[j]][0]
    C=C+[c]

m=min(C)
print('DP Minimum cost:',m)
d={k:p for (k,p) in zip(C,s)}
r=d[m]
r=[x+1 for x in r]
r.insert(0,1)
r.append(1)
print('DPOptimal tour:',r)
return

####    getting inputs from user
n=int(input('Enter the no.of cities:'))
a=[[0 for i in range(n)]for j in range(n)]
print('Enter the cost matrix:')
for i in range(n):
for j in range(n):
    a[i][j]=int(input())
# calling branch & bound
bb(a,n)
#calling dynamic programming
dp(a,n)

Answer 1

基本上就是让别人帮你写申请书。在

对某些人来说很难掌握它，但它应该不会超过几个小时。如果您编写了一个单独的应用程序，在按下按钮时只显示“hello world”，就像您将学习here一样，那么这只是重复代码并使您的逻辑适合的情况。在

当代码没有按照您的想法工作时，社区会很乐意为您提供帮助。试着复习一下教程，然后再回来讨论你的问题。在

附带说明：发布时在代码中添加更多空白/注释，这样会使答案更快、更具相关性。在