0 or 1（最短路转化）

Description
Given a n*n matrix C ij (1<=i,j<=n),We want to find a n*n matrix X ij (1<=i,j<=n),which is 0 or 1.

Besides,X ij meets the following conditions:

1.X 12+X 13+…X 1n=1
2.X 1n+X 2n+…X n-1n=1
3.for each i (1

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<vector>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
int a[305][305],d[305],n,f[305];
bool leap[305];
void spfa(int x)
{
    for(int i = 1;i <= n;i++)leap[i] = false;
    queue<int>p;
    for(int i = 1;i <= n;i++)
    {
        if(i == x)
            d[i] = inf;
        else
        {
            d[i] = a[x][i];
            p.push(i);
            leap[i] = true;
        }
    }
    while(p.size())
    {
        int u = p.front();p.pop();
        leap[u] = false;
        for(int i = 1;i <= n;i++)
        {
            if(d[i] > d[u] + a[u][i])
            {
                d[i] = d[u] + a[u][i];
                if(!leap[i])
                {
                    leap[i] = true;p.push(i);
                }
            }
        }
    }
}
int main()
{
    #ifdef LOCAL
    freopen("C:\\Users\\巍巍\\Desktop\\in.txt","r",stdin);
    //freopen("C:\\Users\\巍巍\\Desktop\\out.txt","w",stdout);
    #endif // LOCAL
    while(scanf("%d",&n) != EOF)
    {
        for(int i = 1;i <= n;i++)
            for(int j = 1;j <= n;j++)
            scanf("%d",&a[i][j]);
        int a,b;
        spfa(1);
        a = d[1];b = d[n];
        //cout<<a<<" "<<b<<endl;
        spfa(n);
        //cout<<d[1]<<" "<<d[n]<<endl;
        int ans = min(b,a + d[n]);
        printf("%d\n",ans);
    }
    return 0;
}