[hdu5402 Travelling Salesman Problem]YY

题意：给一个n*m的矩形，每个格子有一个非负数，求一条从(1,1)到(n,m)的路径（不能经过重复的格子），使得经过的数的和最大，输出具体的方案

思路：对于row为奇数的情况，一行行扫下来即可全部走完得到最大和，对于col为奇数的情况一列列扫即可。对于行和列全部为偶数的情况，将所有格子进行黑白染色，起点和终点的颜色一样，而路径上的颜色是交替的，说明总有一个点不能走到，枚举得到不可到点上的最小值，总和减去就是答案。具体的方案构造方法如下：由于只有一个格子被挖掉不能走，考虑整行或整列的走，走完这个格子前面的所有格子，然后把后面的两行或两列走完，这两行或两列相当于一行或一列，那么整个图相当于是奇数行或奇数列的图了，往后走一定可以遍历完。

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#pragma comment(linker, "/STACK:10240000") #include <map> #include <set> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <vector> #include <cstdio> #include <string> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define X first #define Y second #define pb push_back #define mp make_pair #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) #define copy(a, b) memcpy(a, b, sizeof(a)) typedef long long ll; typedef pair<int, int> pii; typedef unsigned long long ull; //#ifndef ONLINE_JUDGE void RI(vector<int>&a,int n){a.resize(n);for(int i=;i<n;i++)scanf("%d",&a[i]);} void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?:-; while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> void print(const T t){cout<<t<<endl;}template<typename F,typename...R> void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> void print(T*p, T*q){int d=p<q?:-;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} //#endif template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} const double PI = acos(-1.0); const int INF = 1e9 + ; const double EPS = 1e-12; /* -------------------------------------------------------------------------------- */ int n, m, sum; int a[][]; void out() { printf("%d\n", sum); if (n & ) { char ch = 'R'; for (int i = ; i < n; i ++) { for (int j = ; j < m; j ++) putchar(ch); if (i < n - ) putchar('D'); ch = ch == 'L'? 'R' : 'L'; } } else { char ch = 'D'; for (int j = ; j < m; j ++) { for (int i = ; i < n; i ++) putchar(ch); if (j < m - ) putchar('R'); ch = ch == 'D'? 'U' : 'D'; } } putchar('\n'); } void work() { int minnum = INF, x, y; for (int i = ; i < n; i ++) { for (int j = ; j < m; j ++) { bool r = i & , c = j & ; if ((r == c)) continue; if (umin(minnum, a[i][j])) { x = i; y = j; } } } printf("%d\n", sum - minnum); if (x & ) { char ch = 'D'; for (int j = ; j < y; j ++) { for (int i = ; i < n; i ++) putchar(ch); putchar('R'); ch = ch == 'D'? 'U' : 'D'; } ch = 'R'; for (int i = ; i < x; i ++) { putchar(ch); putchar('D'); ch = ch == 'L'? 'R' : 'L'; } for (int i = x + ; i < n; i ++) { putchar('D'); putchar(ch); ch = ch == 'L'? 'R' : 'L'; } if (y < m - ) { putchar('R'); ch = 'U'; for (int j = y + ; j < m; j ++) { for (int i = ; i < n; i ++) putchar(ch); if (j < m - ) putchar('R'); ch = ch == 'D'? 'U' : 'D'; } } } else { char ch = 'R'; for (int i = ; i < x; i ++) { for (int j = ; j < m; j ++) putchar(ch); putchar('D'); ch = ch == 'R'? 'L' : 'R'; } ch = 'D'; for (int j = ; j < y; j ++) { putchar(ch); putchar('R'); ch = ch == 'U'? 'D' : 'U'; } for (int j = y + ; j < m; j ++) { putchar('R'); putchar(ch); ch = ch == 'U'? 'D' : 'U'; } if (x < n - ) { putchar('D'); ch = 'L'; for (int i = x + ; i < n; i ++) { for (int j = ; j < m; j ++) putchar(ch); if (i < n - ) putchar('D'); ch = ch == 'R'? 'L' : 'R'; } } } putchar('\n'); } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE while (cin >> n >> m) { sum = ; for (int i = ; i < n; i ++) { for (int j = ; j < m; j ++) { scanf("%d", &a[i][j]); sum += a[i][j]; } } if (n % || m % ) out(); else work(); } return ; }