hdu5422#53-div2-1Rikka with Graph

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?

当我理解明白题意时，内心几乎是崩溃的==

勇太有一张 $n$ 个点 $m$ 条边的无向图，每一条边的长度都是1。现在他想再在这张图上连上一条连接两个不同顶点边，使得1号点到 $n$ 号点的最短路尽可能的短。现在他想要知道最短路最短是多少，以及再保证最短路最短的情况下，他有多少种连边的方案。

题解：看了提问才明白==最后的最短路一定是1，要是原来1和n没连上，只能连他俩 否则N个点中任意找两个连上即可 亏我还费劲找最短路模板和组合数求法==

#include <iostream>
#include<cstdio>
#include<cstring>
#define MaxN 110
#define MaxInt 65535
using namespace std;
int n,m;
bool A[102][102];
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1; i<=n; i++)
        {
            for(int j=i+1; j<=n; j++)
                A[i][j]=A[j][i]=0;
        }
        int a,b;
        while(m--)
        {
            scanf("%d%d",&a,&b);
            A[b][a]=A[a][b]=1;
        }
        printf("1 ");
       /* start=1;nd=n;
        int tmp=dijkstra(),tmp1;
        int mark=0;
        int num[10000];
        for(int i=1;i<n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                if(A[i][j]!=MAXINT)
                {
                    A[i][j]=A[j][i]=1;
                    tmp1=dijkstra();
                    if(tmp1<tmp)       num[++mark]=1,tmp=tmp1;
                    else if(tmp1==tmp) num[mark]++;
                    A[i][j]=A[j][i]=MAXINT;
                }
            }
        }*/

        if(!A[1][n]) printf("1\n");
        else printf("%d\n",n*(n-1)/2);
    }
    return 0;
}