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=0;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?1:-1; 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?1:-1;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 + 7; const double EPS = 1e-12; /* -------------------------------------------------------------------------------- */ int n, m, sum; int a[102][102]; void out() { printf("%d\n", sum); if (n & 1) { char ch = ‘R‘; for (int i = 0; i < n; i ++) { for (int j = 1; j < m; j ++) putchar(ch); if (i < n - 1) putchar(‘D‘); ch = ch == ‘L‘? ‘R‘ : ‘L‘; } } else { char ch = ‘D‘; for (int j = 0; j < m; j ++) { for (int i = 1; i < n; i ++) putchar(ch); if (j < m - 1) putchar(‘R‘); ch = ch == ‘D‘? ‘U‘ : ‘D‘; } } putchar(‘\n‘); } void work() { int minnum = INF, x, y; for (int i = 0; i < n; i ++) { for (int j = 0; j < m; j ++) { bool r = i & 1, c = j & 1; if ((r == c)) continue; if (umin(minnum, a[i][j])) { x = i; y = j; } } } printf("%d\n", sum - minnum); if (x & 1) { char ch = ‘D‘; for (int j = 0; j < y; j ++) { for (int i = 1; i < n; i ++) putchar(ch); putchar(‘R‘); ch = ch == ‘D‘? ‘U‘ : ‘D‘; } ch = ‘R‘; for (int i = 0; i < x; i ++) { putchar(ch); putchar(‘D‘); ch = ch == ‘L‘? ‘R‘ : ‘L‘; } for (int i = x + 1; i < n; i ++) { putchar(‘D‘); putchar(ch); ch = ch == ‘L‘? ‘R‘ : ‘L‘; } if (y < m - 2) { putchar(‘R‘); ch = ‘U‘; for (int j = y + 2; j < m; j ++) { for (int i = 1; i < n; i ++) putchar(ch); if (j < m - 1) putchar(‘R‘); ch = ch == ‘D‘? ‘U‘ : ‘D‘; } } } else { char ch = ‘R‘; for (int i = 0; i < x; i ++) { for (int j = 1; j < m; j ++) putchar(ch); putchar(‘D‘); ch = ch == ‘R‘? ‘L‘ : ‘R‘; } ch = ‘D‘; for (int j = 0; j < y; j ++) { putchar(ch); putchar(‘R‘); ch = ch == ‘U‘? ‘D‘ : ‘U‘; } for (int j = y + 1; j < m; j ++) { putchar(‘R‘); putchar(ch); ch = ch == ‘U‘? ‘D‘ : ‘U‘; } if (x < n - 2) { putchar(‘D‘); ch = ‘L‘; for (int i = x + 2; i < n; i ++) { for (int j = 1; j < m; j ++) putchar(ch); if (i < n - 1) 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 = 0; for (int i = 0; i < n; i ++) { for (int j = 0; j < m; j ++) { scanf("%d", &a[i][j]); sum += a[i][j]; } } if (n % 2 || m % 2) out(); else work(); } return 0; } |
时间: 2024-10-20 02:48:19