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2013
11-12

POJ 3522 Slim Span [解题报告] Java

Slim Span

问题描述 :

Given an undirected weighted graph G, you should find one of spanning trees specified as follows.

The graph G is an ordered pair (V, E), where V is a set of vertices {v1, v2, …, vn} and E is a set of undirected edges {e1, e2, …, em}. Each edge eE has its weight w(e).

A spanning tree T is a tree (a connected subgraph without cycles) which connects all the n vertices with n − 1 edges. The slimness of a spanning tree T is defined as the difference between the largest weight and the smallest weight among the n − 1 edges of T.


Figure 5: A graph G and the weights of the edges

For example, a graph G in Figure 5(a) has four vertices {v1, v2, v3, v4} and five undirected edges {e1, e2, e3, e4, e5}. The weights of the edges are w(e1) = 3, w(e2) = 5, w(e3) = 6, w(e4) = 6, w(e5) = 7 as shown in Figure 5(b).


Figure 6: Examples of the spanning trees of G

There are several spanning trees for G. Four of them are depicted in Figure 6(a)~(d). The spanning tree Ta in Figure 6(a) has three edges whose weights are 3, 6 and 7. The largest weight is 7 and the smallest weight is 3 so that the slimness of the tree Ta is 4. The slimnesses of spanning trees Tb, Tc and Td shown in Figure 6(b), (c) and (d) are 3, 2 and 1, respectively. You can easily see the slimness of any other spanning tree is greater than or equal to 1, thus the spanning tree Td in Figure 6(d) is one of the slimmest spanning trees whose slimness is 1.

Your job is to write a program that computes the smallest slimness.

输入:

The input consists of multiple datasets, followed by a line containing two zeros separated by a space. Each dataset has the following format.

n m
a1 b1 w1
am bm wm

Every input item in a dataset is a non-negative integer. Items in a line are separated by a space. n is the number of the vertices and m the number of the edges. You can assume 2 ≤ n ≤ 100 and 0 ≤ mn(n − 1)/2. ak and bk (k = 1, …, m) are positive integers less than or equal to n, which represent the two vertices vak and vbk connected by the kth edge ek. wk is a positive integer less than or equal to 10000, which indicates the weight of ek. You can assume that the graph G = (V, E) is simple, that is, there are no self-loops (that connect the same vertex) nor parallel edges (that are two or more edges whose both ends are the same two vertices).

输出:

For each dataset, if the graph has spanning trees, the smallest slimness among them should be printed. Otherwise, −1 should be printed. An output should not contain extra characters.

样例输入:

4 5
1 2 3
1 3 5
1 4 6
2 4 6
3 4 7
4 6
1 2 10
1 3 100
1 4 90
2 3 20
2 4 80
3 4 40
2 1
1 2 1
3 0
3 1
1 2 1
3 3
1 2 2
2 3 5
1 3 6
5 10
1 2 110
1 3 120
1 4 130
1 5 120
2 3 110
2 4 120
2 5 130
3 4 120
3 5 110
4 5 120
5 10
1 2 9384
1 3 887
1 4 2778
1 5 6916
2 3 7794
2 4 8336
2 5 5387
3 4 493
3 5 6650
4 5 1422
5 8
1 2 1
2 3 100
3 4 100
4 5 100
1 5 50
2 5 50
3 5 50
4 1 150
0 0

样例输出:

1
20
0
-1
-1
1
0
1686
50

解题代码:

//* @author: 
import java.io.*;
import java.util.*;
/*求生成树的最大权值与最小权值的最小差:
 *基本思想:先对边排序. 若其在相邻的一组边能生成树,记录最大权值与最小权值的最小差,枚举所有情况
 *如何保证这样求出的是最优解呢,因为边已经按权值进行排序,相邻的一组边生成的树必然是最小的(上界和下界之差最小)
 */
class cin
{
static BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
static int a,c;
static int nextInt() throws IOException
{
   c=in.read();
   a=0;
   while(c==' '||c=='\r'||c=='\n')c=in.read();
   while(c!=' '&&c!='\r'&&c!='\n')
   {
    a=a*10+c-'0';
    c=in.read();
   }
   return a;
}
}

class E
{
int weight,from,to;
void init(int x,int y,int z)
{
   weight=z;
   from=x;
   to=y;
}
}

class Cmp implements Comparator
{
public int compare(Object a,Object b)
{
   int c=((E)a).weight-((E)b).weight;
   if(c< 0)return -1;
   return 1;
}
}

class Set
{
int father[];
int num;
Set(int n)
{
   num=n;
   father=new int[n+1];
   this.clear();
}

void clear()
{
   int i;
   num=father.length-1;
   for(i=1;i<=num;i++)father[i]=i;
}

int findF(int x)
{
   int t=x;
   while(t!=father[t])
   {
    t=father[t];
   }
   while(x!=father[x])
   {
    father[x]=t;
    x=father[x];
   }
   return t;
}

boolean union(int x,int y)
{
   int fx=findF(x),fy=findF(y);
   if(fx==fy)return false;
   father[fy]=fx;
   num--;
   return true;
}
}

class Kruskal
{
E e[];
int numOfE,numOfTree,i,j,numOfD;
Set points;

Kruskal(E a[],int m,int n)
{
   e=a;
   numOfE=m;
   numOfD=n;
   points=new Set(numOfD);
}

int cost()
{
   int dis=100000,h,k=0;
   Arrays.sort(e,new Cmp());
   for(i=0;i< numOfE;i++)    //主要部分代码
   {
    if(points.union(e[i].from,e[i].to))
    {
     if(points.num==1)   //生成一棵树
     {
      k++;
      while(true)
      {
       points.clear();
       for(h=k;h<=i;h++)
       {
        points.union(e[h].from, e[h].to);
       }
       if(points.num==1)k++; //去掉不必要的边
       else break;
      }
      if(dis>e[i].weight-e[k-1].weight) //计算并保留最小差值
       dis=e[i].weight-e[k-1].weight;
     }
    }
   }
   if(dis==100000)dis=-1;
   return dis; 
}

}

public class Main {
    public static void main(String args[]) throws IOException
    {
    int n,m,i;
    E e[];
    while(true)
    {
       n=cin.nextInt();
       m=cin.nextInt();
       if(n==0&&m==0)break;
       e=new E[m];
       for(i=0;i< m;i++)
       {
        e[i]=new E();
        e[i].init(cin.nextInt(), cin.nextInt(), cin.nextInt());
       }
       Kruskal data=new Kruskal(e,m,n);
       System.out.println(data.cost());
    }
    }
 }

  1. for(int i=1; i<=m; i++){
    for(int j=1; j<=n; j++){
    dp = dp [j-1] + 1;
    if(s1.charAt(i-1) == s3.charAt(i+j-1))
    dp = dp[i-1] + 1;
    if(s2.charAt(j-1) == s3.charAt(i+j-1))
    dp = Math.max(dp [j - 1] + 1, dp );
    }
    }
    这里的代码似乎有点问题? dp(i)(j) = dp(i)(j-1) + 1;这个例子System.out.println(ils.isInterleave("aa","dbbca", "aadbbcb"));返回的应该是false

  2. 算法是程序的灵魂,算法分简单和复杂,如果不搞大数据类,程序员了解一下简单点的算法也是可以的,但是会算法的一定要会编程才行,程序员不一定要会算法,利于自己项目需要的可以简单了解。