hdu 5402 Travelling Salesman Problem
运行这些测试数据都对，但就是wa
//4 4
//2 3 3 3//3 3 3 3
//3 3 3 3
//3 3 3 2
//answer:
//43
//DRRURDDLLLDRRR
//
//4 4
//2 1 3 3
//3 3 3 3
//3 3 3 3
//3 3 3 2
//answer:
//43
//DRRURDDLLLDRRR
//
//4 4
//2 3 3 1
//3 3 3 3
//3 3 3 3
//3 3 3 2
//answer:
//43
//DRURDRDLLLDRRR
//
//4 4
//2 3 3 3
//1 3 3 3
//3 3 3 3
//3 3 3 2
//answer:
//43
//RDRURDDLLLDRRR
//
//4 4
//2 3 3 3
//3 3 1 3
//3 3 3 3
//3 3 3 2
//answer:
//43
//DRURRDDLLLDRRR
//
//4 4
//2 3 3 3
//3 3 3 3
//3 1 3 3
//3 3 3 2
//answer:
//43
//RRRDLLLDDRRURD
//
//4 4
//2 3 3 3
//3 3 3 3
//3 3 3 1
//3 3 3 2
//answer:
//43
//RRRDLLLDDRURDR
//
//4 4
//2 3 3 3
//3 3 3 3
//3 3 3 3
//1 3 3 2
//answer:
//43
//RRRDLLLDRDRURD
//
//4 4
//2 3 3 3
//3 3 3 3
//3 3 3 3
//3 3 1 2
//answer:
//43
//RRRDLLLDDRURRD
//
//3 4
//2 3 3 3
//3 3 3 3
//3 3 3 2
//answer:
//34
//RRRDLLLDRRR
//
//4 3
//2 3 3
//3 3 3
//3 3 3
//3 3 2
//answer:
//34
//DDDRUUURDDD
//
//4 4
//2 3 3 3
//3 1 3 3
//3 3 3 3
//3 3 3 2
//answer:
//41
//DRRURDDLLLDRRR
//
//2 4
//2 3 3 3
//3 3 3 2
//answer:
//19
//DRRURD
//
//2 4
//2 1 3 3
//3 3 3 2
//answer:
//19
//DRRURD
//
//2 4
//2 3 3 1
//3 3 3 2
//answer:
//19
//DRURDR
//
//2 4
//2 3 3 3
//1 3 3 2
//answer:
//19
//RDRURD
//
//2 4
//2 3 3 3
//3 3 1 2
//answer:
//19
//DRURRD
//
//思路 如果n或m是奇数的话 肯定可以走完
//如果 都是偶数的话寻找值最小的点
//行遍历RRDLLD 绕过最小点DRRURDD
//黑白块，i+j的奇偶性
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 100;
int main(){
int maps[maxn][maxn];
int sum;
int n, m, x , y ;
while(scanf("%d%d", &n,&m) != EOF){
sum = 0;
x = 1; y = 2;
memset(maps, 100000, sizeof(maps));
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
scanf("%d", &maps[i][j]);
sum += maps[i][j];
if((i+j)%2&&maps[i][j] < maps[x][y]){
x = i; y = j;
}
}
}
if(n%2||m%2){
printf("%d\n", sum);
if(n%2){
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m-1; j++){
if(i&1) printf("R");
else printf("L");
}
if(i==n) printf("\n");
else printf("D");
}
}
else if(m%2){
for(int i = 1; i <= m; i++){
for(int j = 1; j <= n-1; j++){
if(i&1) printf("D");
else printf("U");
}
if(i==m) printf("\n");
else printf("R");
}
}
}
else {
printf("%d\n", sum - maps[x][y]);
if(x%2){
for(int i = 1; i < x; i++){
for(int j = 1; j <= m - 1 ; j++){
if(i%2) printf("R");
else printf("L");
}
printf("D");
}
for(int i = x, j = 1; j < m ; ){
if((i+j)%2==0 && j>=y){
printf("R");
j++;
}
else if((i+j)%2 && j > y){
if(x==i){
printf("D");
i++;
}
else {
printf("U");
i--;
}
}
else if((i+j)%2 && j < y){
printf("R");
j++;
}
else if((i+j)%2==0 && j < y){
if(x==i){
printf("D");
i++;
}
else {
printf("U");
i--;
}
}
if(i==x+1&&j==m&&x+1!=n){ printf("D");}
else if(i == x&&j==m&&x+1!=n) {printf("DD");}
else if(i == x&&j==m&&x+1==n) printf("D");
}
for(int i = x + 2; i <= n; i++){
for(int j = 1; j <= m-1; j++){
if(i%2) printf("L");
else printf("R");
}
if(i!=n)printf("D");
}
}
else {
for(int i = 1; i < x-1; i++){
for(int j = 1; j <= m-1; j++){
if(i%2) printf("R");
else printf("L");
}
printf("D");
}
for(int i = x - 1, j = 1; j < m ;){
if(j > y && (i+j)%2){
if(x-1==i){
printf("D");
i++;
}
else {
printf("U");
i--;
}
}
else if(j>=y&&(i+j)%2==0){
printf("R");
j++;
}
else if((i+j)%2==0&&j<y){
if(x==i) {
printf("U");
i--;
}
else {
printf("D");
i++;
}
}
else if((i+j)%2 && j < y){
printf("R");
j++;
}
if(i == x&&j==m&&x!=n) printf("D");
else if(i == x-1&&j==m&&x!=n) printf("DD");
else if(i == x-1&&j==m&&x==n) printf("D");
}
for(int i = x + 1 ; i <= n; i++){
for(int j = 1; j <= m-1; j++){
if(i%2) printf("L");
else printf("R");
}
if(i!=n)printf("D");
}
}
printf("\n");
}
}
return 0;
}