HDU 5402 Travelling Salesman problem (mut#9 violence simulation)

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"The main idea": Walk the Square, from "" to "n,m", the middle process to get the number and maximum, and the output path

"Ideas":

If there is an odd number in N and M, then all of them will be all right. Otherwise, to find a minimum point, do not, the coordinates of this point to meet x+y is odd if not, discard the point will certainly lead to another point also can not go. Then find the point where the brute force output path can be.

Code:

#include <bits/stdc++.h>using namespace Std;const int n=105;typedef long long LL; LL Mat[n][n]; LL n,m;    LL x,y;void getmin () {///In the case of n,m having an even number, it is necessary to discard a point and find the point with the lowest weight (the sum of coordinates must be odd) x=1,y=2;                for (int i=1, i<=n; ++i) for (int j=1; j<=m; ++j) {if (mat[i][j]<mat[x][y]&& (i+j) &1) {                X=i;            Y=j;        }}}int Main () {while (~scanf ("%i64d%i64d", &n,&m)) {memset (mat,0,sizeof (MAT));        LL sum=0;                for (int. I=1; i<=n; ++i) {for (int j=1; j<=m; ++j) {scanf ("%i64d", &mat[i][j]);   SUM+=MAT[I][J];        Statistics and} LL ans; if (n&1| |            M&AMP;1)////If there is an odd number, the squares can all go through, {printf ("%i64d\n", sum); if (n&1| | (n&1&&m&1)) {for (int i=0, i<n; ++i) {for (int j=1; j<m; ++j) {if (i&                        1) printf ("L");  else printf ("R");                  } if (i!=n-1) printf ("D");            } puts (""); } else if (m&1&&n%2==0) {for (Int. i=0; i<m; ++i) {for (int j=1; j <n;                        ++J) {if (i&1) printf ("U");                    else printf ("D");                } if (i!=m-1) printf ("R");            } puts ("");       }} else{///Even condition getmin ();            The coordinates of the point at which the minimum weight is obtained ans=sum-mat[x][y];            printf ("%i64d\n", ans); for (int i = 1; I <= n; i + = 2) {if (x = = I | | x = = i + 1) {///the path for the point is processed for (int j = 1 ; J < y;                        J + +) {if (J & 1) printf ("D"); else printf ("U");                    printf ("R");                    } if (y<m) printf ("R");             for (int j = y + 1; j <= M; j + +) {           if (J & 1) printf ("U");                        else printf ("D");                    if (J < m) printf ("R");                } if (I < n-1) printf ("D");                    } else if (I < x) {///the path before the point for (int j = 1; j < M; j + +) printf ("R");                    printf ("D");                    for (int j = 1; j < M; j + +) printf ("L");                printf ("D");                    } else {///the path after the point for (int j = 1; j < M; j + +) printf ("L");                    printf ("D");                    for (int j = 1; j < M; j + +) printf ("R");                if (i < n-1) printf ("D");        }} puts (""); }} return 0;}

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HDU 5402 Travelling Salesman problem (mut#9 violence simulation)