hdu 5402 Travelling Salesman Problem

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原来那道模拟题是这样的，因为我走的奇偶决定了我现在是在左还是在右，我刚开始想的是要走到断点的上一层或本层，但实际上应该固定每次只走两行，这样每次都是从左边开始了，然后我只需要考虑从左边去把空格绕了就行，然后就是从右边下来了，然后就反着走就可以了~还是需要加强对这种思维题目的练习~

#pragma comment(linker, "/STACK:1024000000,1024000000") #include<set>#include<map>#include<queue>#include<stack>#include<cmath>#include<cstdio>#include<vector>#include<string>#include<cstring>#include<algorithm>#include<functional>using namespace std;typedef long long LL;const LL base = 1e9 + 7;const int maxn = 105;LL T, n, m, a[maxn][maxn], sum, x, y;inline void read(int &x){    char ch;    while ((ch = getchar())<'0' || ch>'9');    x = ch - '0';    while ((ch = getchar()) >= '0' && ch <= '9') x = x * 10 + ch - '0';}void get(){    x = 1;    y = 2;    for (int i = 1; i <= n;i++)        for (int j = 1; j <= m; j++)             if (((i + j) & 1) && a[x][y] > a[i][j]) x = i, y = j;}int main(){    while (scanf("%lld%lld", &n, &m) !=EOF)    {        sum = 0;        for (int i = 1; i <= n;i++)            for (int j = 1; j <= m; j++)            {                scanf("%lld", &a[i][j]);                sum += a[i][j];            }        if (n & 1 || m & 1)        {            printf("%lld\n", sum);            if (n & 1)            {                for (int i = 1; i <= n; i++)                {                    for (int j = 1; j < m; j++)                         if (i & 1) printf("R"); else printf("L");                    if (i < n) printf("D"); else printf("\n");                }            }            else            {                for (int i = 1; i <= m; i++)                {                    for (int j = 1; j < n; j++)                        if (i & 1) printf("D"); else printf("U");                    if (i < m) printf("R"); else printf("\n");                }            }        }        else        {            get();            printf("%lld\n", sum - a[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");                }                else if (i < x)                {                    for (int j = 1; j < m; j++) printf("R");                    printf("D");                    for (int j = 1; j < m; j++) printf("L");                    printf("D");                }                else                {                    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");                }            }            printf("\n");        }    }    return 0;}

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