• [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=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
", 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('
');
}

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
", 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('
');
}

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;
}
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  • 原文地址：https://www.cnblogs.com/jklongint/p/4741132.html
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