hdu5402:Travelling Salesman Problem

Posted 2020-10-05 ck666

　Travelling Salesman Problem

 

2017-09-01

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Problem Description

Teacher Mai is in a maze with n rows and m columns. There is a non-negative number in each cell. Teacher Mai wants to walk from the top left corner (1,1) to the bottom right corner (n,m). He can choose one direction and walk to this adjacent cell. However, he can‘t go out of the maze, and he can‘t visit a cell more than once.
Teacher Mai wants to maximize the sum of numbers in his path. And you need to print this path.

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Input

There are multiple test cases.
For each test case, the first line contains two numbers n,m(1≤n,m≤100,n?m≥2).
In following n lines, each line contains m numbers. The j-th number in the i-th line means the number in the cell (i,j). Every number in the cell is not more than 104.

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Output

For each test case, in the first line, you should print the maximum sum.
In the next line you should print a string consisting of "L","R","U" and "D", which represents the path you find. If you are in the cell (x,y), "L" means you walk to cell(x,y?1), "R" means you walk to cell (x,y+1), "U" means you walk to cell (x?1,y), "D" means you walk to cell (x+1,y).

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Sample Input

INPUT_1

3 3

2 3 3

3 3 3

3 3 2

INPUT_2

2 2

1 0

1 1

INPUT_3

4 4

6 0 6 6

6 6 6 6

6 6 6 6

6 6 6 6

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Sample Output

OUT_1

25

RRDLLDRR

OUT_2

3

DR

OUT_3

90

DRRURDDLLLDRRR

DDDRUURURDDLDR皆可。。

spj.....

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此题是一个好大的模拟,出题人你节操被10万买走了吧

题意:给你一个大矩阵,一笔从(1,1)走到(n,m)路径上最大的和.

看出来结论:如果n||m有一个是奇数，那么就可以走所有的点,怎么走看心情

n和m全为偶数就找到一个i,j(1<=i<n;1<=j<=m)&&(i+j)是奇数的点，是绕过这个看不顺的点走到(n,m)

路程全部模拟;

要不是掉rating我才不干呢,什么破玩意

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int read(){
    char ch=getchar();
    int f=1,an=0;
    while(!(‘0‘<=ch&&ch<=‘9‘)){if(ch==‘-‘)f=-f;ch=getchar();}
    while(‘0‘<=ch&&ch<=‘9‘){an=an*10+(ch-‘0‘);ch=getchar();}
    return an*f;
}
int n,m,sum;
int x,y,z;
int a[100+99][100+99];
void P1(int n,int m){
    cout<<sum<<endl;
    for(int i=1;i<=n;i++){
        if(i&1){
            for(int j=1;j<m;j++)cout<<"R";if(i!=n)cout<<"D";}
        else {
            for(int j=1;j<m;j++)cout<<"L";if(i!=n)cout<<"D";}
    }
    cout<<endl;
}
void P2(int n,int m){
    cout<<sum<<endl;
    for(int i=1;i<=m;i++){
        if(i&1){
            for(int j=1;j<n;j++)cout<<"D";if(i!=m)cout<<"R";}
        else{
            for(int j=1;j<n;j++)cout<<"U";if(i!=m)cout<<"R";}
    }
    cout<<endl;
}
void P5(int a,int b,int c,int d){
    int e=a;
    e=(e-1)>>1;
    for(int i=1;i<=e;i++){
        for(int j=1;j<d;j++)cout<<"R";cout<<"D";
        for(int j=1;j<d;j++)cout<<"L";cout<<"D";}
    int k=1,z=1;bool flag=1;
    while(z<d){
        if(z==b){cout<<"R";z++;}
        else{
            if(flag){cout<<"DR";flag^=1;z++;}
            else {cout<<"UR";flag^=1;z++;}
        }
    }
    if(b!=m)cout<<"D";
    c>>=1;c-=e;
    for(int i=2;i<=c;i++){
        if(i==2)cout<<"D";
        for(int j=1;j<d;j++)cout<<"L";cout<<"D";
        for(int j=1;j<d;j++)cout<<"R";if(i!=c)cout<<"D";}
}
void P3(int n,int m){
    cout<<sum-a[x][y]<<endl;
    P5(x,y,n,m);
    cout<<endl;
}
void G(){sum=0;n=m=0;}
int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        z=9999999;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++){a[i][j]=read();sum+=a[i][j];
            if((i+j)&1)if(z>a[i][j]){z=a[i][j];x=i;y=j;}
        }
        if(n&1)P1(n,m);
        else if(m&1)P2(n,m);
        else P3(n,m);
        G();
    }
    return 0;
}

有节操模拟

by:s_a_b_e_r

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 什么破玩意+1

题意见上边（实力甩锅

n或者m任意一个是奇数的话就一笔画走完这个矩形

两个都是偶数的话就麻烦了x

要在矩阵上去掉一个点

但是不是所有点都能去

4x4的点图如下

A X C X
X C X C
C X C X
X C X B

其中X是可以去掉的点，C是不能去掉的点

关于怎么去点……这个做法很多啊w

表示可以每次扫两行

两行内都没有要去掉的点的话就直接走直线

有的话就扭过去x

大概这样

  →      →

↑  ↓→↑     ……

……我自己都看不懂了x

……没写完下午再补