2017 ACM-ICPC 亚洲区(南宁赛区)网络赛 Minimum Distance in a Star Graph
2017-09-24 20:06
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In this problem, we will define a graph called star graph, and the question is to find the minimum distance between two given nodes in the star graph.
Given an integer nn,
an n-dimensionaln−dimensionalstar
graph, also referred to as S_{n}Sn,
is an undirected graph consisting of n!n! nodes
(or vertices) and ((n-1)\
*\ n!)/2((n−1) ∗ n!)/2 edges.
Each node is uniquely assigned a label x_{1}\
x_{2}\ ...\ x_{n}x1 x2 ... xnwhich
is any permutation of the n digits {1,
2, 3, ..., n}1,2,3,...,n.
For instance, an S_{4}S4 has
the following 24 nodes {1234,
1243, 1324, 1342, 1423, 1432, 2134, 2143, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 3412, 3421, 4123, 4132, 4213, 4231, 4312, 4321}1234,1243,1324,1342,1423,1432,2134,2143,2314,2341,2413,2431,3124,3142,3214,3241,3412,3421,4123,4132,4213,4231,4312,4321.
For each node with label x_{1}\
x_{2} x_{3}\ x_{4}\ ...\ x_{n}x1 x2x3 x4 ... xn,
it has n-1n−1 edges
connecting to nodes x_{2}\
x_{1}\ x_{3}\ x_{4}\ ...\ x_{n}x2 x1 x3 x4 ... xn, x_{3}\
x_{2}\ x_{1}\ x_{4}\ ...\ x_{n}x3 x2 x1 x4 ... xn, x_{4}\
x_{2}\ x_{3}\ x_{1}\ ...\ x_{n}x4 x2 x3 x1 ... xn,
..., and x_{n}\
x_{2}\ x_{3}\ x_{4}\ ...\ x_{1}xn x2 x3 x4 ... x1.
That is, the n-1n−1 adjacent
nodes are obtained by swapping the first symbol and the d-thd−th symbol
of x_{1}\
x_{2}\ x_{3}\ x_{4}\ ...\ x_{n}x1 x2 x3 x4 ... xn,
for d
= 2, ..., nd=2,...,n.
For instance, in S_{4}S4,
node 12341234has 33 edges
connecting to nodes 21342134, 32143214,
and 42314231.
The following figure shows how S_{4}S4looks
(note that the symbols aa, bb, cc,
and dd are
not nodes; we only use them to show the connectivity between nodes; this is for the clarity of the figure).
![](https://res.jisuanke.com/img/upload/20170920/b9510b5515aa0293915340ea604adac52bc73f30.png)
In this problem, you are given the following inputs:
nn:
the dimension of the star graph. We assume that nn ranges
from 44 to 99.
Two nodes x_{1}x1 x_{2}x2 x_{3}x3 ... x_{n}xn and y_{1}y1 y_{2}y2 y_{3}\
...\ y_{n}y3 ... yn in S_{n}Sn.
You have to calculate the distance between these two nodes (which is an integer).
nn (dimension
of the star graph)
A list of 55 pairs
of nodes.
A list of 55 values,
each representing the distance of a pair of nodes.
2017
ACM-ICPC 亚洲区(南宁赛区)网络赛
题意:给你1-N的N个数,构建N维(几维其实不重要),然后构图的规则是相邻的顶点只能是由第一个字符跟2-n的字符进行交换得来的,问你给出的五个询问中两个节点的最短距离分别是多少?
思路:宽度优先搜索加剪枝,对搜过的节点进行弹出
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string.h>
#include<string>
#include<set>
#include<queue>
using namespace std;
string numa;
string numb;
int n;
struct node{
string str;
int times;
};
int bfs() {
queue<node> q;
set<string> ss;
node now,next;
now.str = numa;
now.times = 0;
q.push(now);
ss.insert(now.str);
while (!q.empty()) {
now = q.front();
q.pop();
if (now.str == numb)
return now.times;
for (int i = 1; i < n; ++i) {
next.str = now.str;
next.times=now.times+1;
next.str[i] = now.str[0];
next.str[0] = now.str[i];
if (ss.count(next.str) == 0)
{
ss.insert(next.str);
q.push(next);
}else
continue;
}
}
}
int main()
{
scanf("%d", &n);
int t = 5;
while (t--) {
cin >> numa >> numb;
cout << bfs() << endl;
}
return 0;
}
Given an integer nn,
an n-dimensionaln−dimensionalstar
graph, also referred to as S_{n}Sn,
is an undirected graph consisting of n!n! nodes
(or vertices) and ((n-1)\
*\ n!)/2((n−1) ∗ n!)/2 edges.
Each node is uniquely assigned a label x_{1}\
x_{2}\ ...\ x_{n}x1 x2 ... xnwhich
is any permutation of the n digits {1,
2, 3, ..., n}1,2,3,...,n.
For instance, an S_{4}S4 has
the following 24 nodes {1234,
1243, 1324, 1342, 1423, 1432, 2134, 2143, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 3412, 3421, 4123, 4132, 4213, 4231, 4312, 4321}1234,1243,1324,1342,1423,1432,2134,2143,2314,2341,2413,2431,3124,3142,3214,3241,3412,3421,4123,4132,4213,4231,4312,4321.
For each node with label x_{1}\
x_{2} x_{3}\ x_{4}\ ...\ x_{n}x1 x2x3 x4 ... xn,
it has n-1n−1 edges
connecting to nodes x_{2}\
x_{1}\ x_{3}\ x_{4}\ ...\ x_{n}x2 x1 x3 x4 ... xn, x_{3}\
x_{2}\ x_{1}\ x_{4}\ ...\ x_{n}x3 x2 x1 x4 ... xn, x_{4}\
x_{2}\ x_{3}\ x_{1}\ ...\ x_{n}x4 x2 x3 x1 ... xn,
..., and x_{n}\
x_{2}\ x_{3}\ x_{4}\ ...\ x_{1}xn x2 x3 x4 ... x1.
That is, the n-1n−1 adjacent
nodes are obtained by swapping the first symbol and the d-thd−th symbol
of x_{1}\
x_{2}\ x_{3}\ x_{4}\ ...\ x_{n}x1 x2 x3 x4 ... xn,
for d
= 2, ..., nd=2,...,n.
For instance, in S_{4}S4,
node 12341234has 33 edges
connecting to nodes 21342134, 32143214,
and 42314231.
The following figure shows how S_{4}S4looks
(note that the symbols aa, bb, cc,
and dd are
not nodes; we only use them to show the connectivity between nodes; this is for the clarity of the figure).
![](https://res.jisuanke.com/img/upload/20170920/b9510b5515aa0293915340ea604adac52bc73f30.png)
In this problem, you are given the following inputs:
nn:
the dimension of the star graph. We assume that nn ranges
from 44 to 99.
Two nodes x_{1}x1 x_{2}x2 x_{3}x3 ... x_{n}xn and y_{1}y1 y_{2}y2 y_{3}\
...\ y_{n}y3 ... yn in S_{n}Sn.
You have to calculate the distance between these two nodes (which is an integer).
Input Format
nn (dimensionof the star graph)
A list of 55 pairs
of nodes.
Output Format
A list of 55 values,each representing the distance of a pair of nodes.
样例输入
4 1234 4231 1234 3124 2341 1324 3214 4213 3214 2143
样例输出
1 2 2 1 3
题目来源
2017ACM-ICPC 亚洲区(南宁赛区)网络赛
题意:给你1-N的N个数,构建N维(几维其实不重要),然后构图的规则是相邻的顶点只能是由第一个字符跟2-n的字符进行交换得来的,问你给出的五个询问中两个节点的最短距离分别是多少?
思路:宽度优先搜索加剪枝,对搜过的节点进行弹出
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string.h>
#include<string>
#include<set>
#include<queue>
using namespace std;
string numa;
string numb;
int n;
struct node{
string str;
int times;
};
int bfs() {
queue<node> q;
set<string> ss;
node now,next;
now.str = numa;
now.times = 0;
q.push(now);
ss.insert(now.str);
while (!q.empty()) {
now = q.front();
q.pop();
if (now.str == numb)
return now.times;
for (int i = 1; i < n; ++i) {
next.str = now.str;
next.times=now.times+1;
next.str[i] = now.str[0];
next.str[0] = now.str[i];
if (ss.count(next.str) == 0)
{
ss.insert(next.str);
q.push(next);
}else
continue;
}
}
}
int main()
{
scanf("%d", &n);
int t = 5;
while (t--) {
cin >> numa >> numb;
cout << bfs() << endl;
}
return 0;
}
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