HDU 5402 强模拟

HDU 5402

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

给一个n*m矩阵，矩阵每个格子有权值。问怎样不重复的走能使得权值和最大。

思路：

大模拟。

n与m中有一个奇数，就可以全部走完。

n与m都为偶数，需要舍弃一个权值最小且满足(i + j) % 2 != 0的点。

具体就自己模拟去吧，我也WA了无数发，改出来的感觉真好。

源码：

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <cmath>

#include <string>

#include <algorithm>

#include <iostream>

#include <queue>

using namespace std;

#define LL long long

const int MAXN = 100 + 5;

int a[MAXN][MAXN];

int n, m;

void Ytravel(int j)

{

if(j % 2 == 0)

for(int k = 2 ; k <= n ; k++)

printf("U");

else

for(int k = 2 ; k <= n ; k++)

printf("D");

}

void Xtravel(int i)

{

if(i % 2 == 0)

for(int k = 2 ; k <= m ; k++)

printf("L");

else

for(int k = 2 ; k <= m ; k++)

printf("R");

}

void XXtravel(int i, int x, int y)

{

// printf("XX\n");

int j, dj;

if(i % 2 == 0){

j = m, dj = -1;

}

else

j = 1, dj = 1;

int flag = !(i % 2);

while(j > 0 && j <= m){

if(j == y){

flag = !flag;

}

else if(flag == 0)

printf("D");

else if(flag == 1)

printf("U");

if(dj == -1 && j != 1)

printf("L");

else if(dj == 1 && j != m)

printf("R");

// printf("\nj = %d\n", j);

j += dj;

flag = !flag;

}

}

void YYtravel(int i, int x, int y)

{

// printf("YY\n");

// printf("x = %d, y = %d\n", x, y);

int j, dj;

if(i % 2 == 0){

j = n, dj = -1;

}

else

j = 1, dj = 1;

int flag = !(i % 2);

while(j > 0 && j <= n){

if(j == x){

flag = !flag;

}

else if(flag == 0)

printf("R");

else if(flag == 1)

printf("L");

if(dj == -1 && j != 1)

printf("U");

else if(dj == 1 && j != n)

printf("D");

j += dj;

flag = !flag;

}

}

int main()

{

while(scanf("%d%d", &n, &m) != EOF){

LL sum = 0;

int mmin = 10000 + 5;

int x, y;

for(int i = 1 ; i <= n ; i++)

for(int j = 1 ; j <= m ; j++){

scanf("%d", &a[i][j]);

sum += a[i][j];

if((i + j) % 2 != 0){

if(mmin > a[i][j]){

mmin = a[i][j];

x = i, y = j;

}

}

}

if(n % 2 != 0){

printf("%I64d\n", sum);

for(int i = 1 ; i <= n ; i++){

Xtravel(i);

if(i != n)

printf("D");

}

printf("\n");

}

else if(m % 2 != 0){

printf("%I64d\n", sum);

for(int i = 1 ; i <= m ; i++){

Ytravel(i);

if(i != m)

printf("R");

}

printf("\n");

}

else{

printf("%I64d\n", sum - mmin);

int i, j;

if(y != 1 && y != m){

int flag = 0;

for(i = 1 ; i <= n ; i++){

if((i-1)/2 != (x-1)/2){

Xtravel(i - flag);

}

else{

XXtravel(i++, x, y);

flag = 1;

}

if(i < n)

printf("D");

}

}

else{

int flag = 0;

for(i = 1 ; i <= m ; i++){

if((i-1)/2 != (y-1)/2){

Ytravel(i - flag);

}

else{

flag = 1;

YYtravel(i++, x, y);

}

if(i < m)

printf("R");

}

}

printf("\n");

}

}

return 0;

}

/*

4 4

1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4

4 4

1 1 1 1

2 2 2 2

3 3 3 3

4 4 4 4

2 2

1 2

3 4

2 2

1 3

2 4

2 4

3 3 3 1

3 3 3 3

*/