Rikka with Graph
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 182 Accepted Submission(s): 95
Problem Description
As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:
Yuta has a non-direct graph with n
vertices and m
edges. The length of each edge is 1. Now he wants to add exactly an edge which connects two different vertices and minimize the length of the shortest path between vertice 1 and vertice n
. Now he wants to know the minimal length of the shortest path and the number of the ways of adding this edge.
It is too difficult for Rikka. Can you help her?
Yuta has a non-direct graph with n
vertices and m
edges. The length of each edge is 1. Now he wants to add exactly an edge which connects two different vertices and minimize the length of the shortest path between vertice 1 and vertice n
. Now he wants to know the minimal length of the shortest path and the number of the ways of adding this edge.
It is too difficult for Rikka. Can you help her?
Input
There are no more than 100 testcases.
For each testcase, the first line contains two numbers n,m(2≤n≤100,0≤m≤100)
.
Then m
lines follow. Each line contains two numbers u,v(1≤u,v≤n)
, which means there is an edge between u
and v
. There may be multiedges and self loops.
For each testcase, the first line contains two numbers n,m(2≤n≤100,0≤m≤100)
.
Then m
lines follow. Each line contains two numbers u,v(1≤u,v≤n)
, which means there is an edge between u
and v
. There may be multiedges and self loops.
Output
For each testcase, print a single line contains two numbers: The length of the shortest path between vertice 1 and vertice n
and the number of the ways of adding this edge.
and the number of the ways of adding this edge.
Sample Input
2 1
1 2
Sample Output
1 1
这道题看似是一个图论题， 然而仔细一看就水的不行啦。 如果 1 和点 n 没有直接相连的路径， 那么最短路径就是这一条。 否则，最短路径还是这一条。 但是连线的方案确是任意乱连都可以。有 n*（n-1）/2个连法。
当时对题意理解错啦： 理解为添加一条路径使1到其他点路径的总和最小。 （呜呜呜呜~~~~）
#include<iostream>
#include<cstdio>
using namespace std; int main()
{
int m, n;
while(~scanf("%d%d", &n, &m))
{
int u, v;
int ok = ;
for(int i=; i<m; i++)
{
scanf("%d%d", &u, &v);
if((u==&&v==n)||(v==&&u==n))
ok = ;
}
if(!ok) printf("%d %d\n",, );
else printf("%d %d\n",, n*(n-)/);
}
return ;
}