HDU5402 Travelling Salesman Problem 【模拟】

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5402

题意：见 https://vjudge.net/problem/HDU-5402

题解：
如果 n m 其中有一个奇数，那么一定可以到达，输出就可以了。
否则 我们不妨以(1,1) (n,m) 为黑色，将该棋盘黑白格染色，我们可以发现，必定有一个白格子没有走到，我们让最小权值白格子不走到即可
比较麻烦，实现过程细节很多，详见代码。
这题没看着多组数据WA了4次，多组数据 n,m 判奇数后直接return 0 WA了1次。。

代码：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <map> // STL#include <string> #include <vector>#include <queue>#include <stack>#define mpr make_pair#define debug() puts("okkkkkkkk")using namespace std;typedef long long LL;const int inf = 1 << 26;int n, m, sum = 0, tmpmx = inf, x, y;int matr[2005][2005];void init() { tmpmx = inf; x = 0; y = 0; sum = 0; memset(matr, 0, sizeof(matr)); }int main(){//    freopen("ACM-A.in", "r", stdin);    while( scanf("%d %d", &n, &m) != EOF ) {        init();        for ( int i = 1; i <= n; i ++ ) {            for ( int j = 1; j <= m; j ++ ) {                scanf("%d", &matr[i][j]);                sum += matr[i][j];                if((i+j)&1 && matr[i][j] < tmpmx) { tmpmx = matr[i][j]; x = i; y = j; }            }        }        if( n&1 || m&1 ) {            printf("%d\n", sum);            if( n&1 ) {                for ( int i = 1; i < n; i += 2) {                    for ( int j = 1; j < m; j ++ ) printf("R");                    printf("D");                    for ( int j = 1; j < m; j ++ ) printf("L");                    printf("D");                }                for ( int j = 1; j < m; j ++ ) printf("R");            } else {                for ( int i = 1; i < m; i += 2) {                     for ( int j = 1; j < n; j ++ ) printf("D");                    printf("R");                    for ( int j = 1; j < n; j ++ ) printf("U");                    printf("R");                }                for ( int j = 1; j < n; j ++ ) printf("D");            }            puts("");            continue;        }         int fg = 1;        printf("%d\n", sum-matr[x][y]);        for ( int i = 1; i <= n; i += 2 ) {            if(x == i || x == i+1 ) {                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");                 fg = 0;            } else if(fg) {                for ( int j = 1; j < m; j ++ ) printf("R");                printf("D");                for ( int j = 1; j < m; j ++ ) printf("L");                printf("D");            } else if(!fg) {                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;}