UVALive 4256 Salesmen （树DP，4级）_uvalive salesman-优快云博客

本文链接：https://blog.youkuaiyun.com/nealgavin/article/details/10456781

本文探讨了一个旅行推销员问题的变体，即为销售员在工作区域内的移动路径寻找最优路径，该路径长度与原始路径相同，但满足特定的连通性和连续性条件。通过构建图模型并应用动态规划算法，实现路径优化。

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B - Salesmen

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Description

Traveling salesmen of nhn. (the prestigious Korean internet company) report their current location to the company on a regular basis. They also have to report their new location to the company if they are moving to another location. The company keep each salesman's working path on a map of his working area and uses this path information for the planning of the next work of the salesman. The map of a salesman's working area is represented as a connected and undirected graph, where vertices represent the possible locations of the salesman an edges correspond to the possible movements between locations. Therefore the salesman's working path can be denoted by a sequence of vertices in the graph. Since each salesman reports his position regularly an he can stay at some place for a very long time, the same vertices of the graph can appear consecutively in his working path. Let a salesman's working path be correct if two consecutive vertices correspond either the same vertex or two adjacent vertices in the graph.

For example on the following graph representing the working area of a salesman,

$\epsfbox{p4256.eps}$

a reported working path [1 2 2 6 5 5 5 7 4] is a correct path. But a reported working path [1 2 2 7 5 5 5 7 4] is not a correct path since there is no edge in the graph between vertices 2 a 7. If we assume that the salesman reports his location every time when he has to report his location (but possibly incorrectly), then the correct path could be [1 2 2 4 5 5 5 7 4], [1 2 4 7 5 5 5 7 4], or [1 2 2 6 5 5 5 7 4].

The length of a working path is the number of vertices in the path. We define the distance between two paths A = a₁a₂...a_n and B = b₁b₂...b_n of the same length n as

dist( A, B) = $\displaystyle \sum^{{n}}_{{i=1}}$ d ( a _i, b _i)

where

d ( a, b) = $\displaystyle \left\{\vphantom{ \begin{array}{cc} 0 & \mbox{if } a=b \\ 1 & \mbox{otherwise} \end{array} }\right.$ $\displaystyle \begin{array}{cc} 0 & \mbox{if } a=b \\ 1 & \mbox{otherwise} \end{array}$

Given a graph representing the working area of a salesman and a working path (possible not a correct path), A , of a salesman, write a program to compute a correct working path, B , of the same length where the distance dist(A, B) is minimized.

Input

The program is to read the input from standard input. The input consists of T test cases. The number of test cases (T) is given in the first line of the input. The first line of each test case contains two integers n₁ , n₂ (3 $\le$ n₁ $\le$ 100, 2 $\le$ n₂ $\le$ 4, 950) where n₁ is the number of vertices of the graph representing the working map of a salesman and n₂ is the number of edges in the graph. The input graph is a connected graph. Each vertex of the graph is numbered from 1 to n₁ . In the following n₂ lines, each line contains a pair of vertices which represent an edge of the graph. The last line of each test case contains information on a working path of the salesman. The first integer n (2 $\le$ n $\le$ 200) in the line is the length of the path and the following n integers represent the sequence of vertices in the working path.

Output

Your program is to write to standard output. Print one line for each test case. The line should contain the minimum distance of the input path to a correct path of the same length.

Sample Input

2 
7 9 
1 2 
2 3 
2 4 
2 6 
3 4 
4 5 
5 6 
7 4 
7 5 
9 1 2 2 7 5 5 5 7 4 
7 9 
1 2 
2 3 
2 4 
2 6 
3 4 
4 5 
5 6 
7 4 
7 5 
9 1 2 2 6 5 5 5 7 4

Sample Output

1 
0

问最少改几个数使得相邻要么相等，要么是树上节点。

思路：建树，添加自身环节点，这就省去了数组中数字相等的判断，直接判断是不是在树上就好了，然后一个DP也

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
const int mm=222;
int dp[mm][mm];
bool g[mm][mm];
int f[mm],n,m;
int DP(int t,int fa)
{ if(t==0)return 0;
  if(dp[t][fa]!=-1)return dp[t][fa];
  int&ret=dp[t][fa];ret=mm;
  FOR(i,1,n)
  {
    if(g[i][fa])
    {
      ret=min(ret,DP(t-1,i)+(i==f[t]?0:1));
    }
  }
  return ret;
}
int main()
{
  int cas,a,b;
  while(~scanf("%d",&cas))
  {
    while(cas--)
    {
      scanf("%d%d",&n,&m);
      clr(g,0);clr(dp,-1);
      FOR(i,1,m)
      {
        scanf("%d%d",&a,&b);
        g[a][b]=1;g[b][a]=1;

      }
      FOR(i,1,n)
      {
        g[i][0]=1;g[0][i]=1;//初始化
       g[i][i]=1;//equal
      }
      scanf("%d",&m);
      FOR(i,1,m)scanf("%d",&f[i]);
      printf("%d\n",DP(m,0));
    }
  }
  return 0;
}