HDU 4370 0 or 1（最短路）

思路：一个很裸的0/1规划的题，没接触过的话很难会朝图论的方面去想，从前两个条件其实就是点1的出度为1，点n的入度为1，然后其他点的出度等于入度，边权就是它的费用，那么建图然后最短路就可以了，这里比较坑的是由于只说了出度为1，但入度不知道，所以可能存在两个环，从1出发回到1，从n出发回到n的情况

#include<bits/stdc++.h>
using namespace std;
#define inf 1e9
const int maxn = 400;
int n,mp[maxn][maxn];
int d[maxn],inq[maxn];
int spfa(int s,int t,int &res)
{
queue<int>q;
for(int i = 0;i<=n;i++)
d[i]=inf;
memset(inq,0,sizeof(inq));
d[s]=0;
q.push(s);
while(!q.empty())
{
int u = q.front();
q.pop();
inq[u]=0;
for(int i = 0;i<n;i++)
{
if(s!=u && i==s)
res = min(res,mp[u][i]+d[u]);
if(d[i]>d[u]+mp[u][i])
{
d[i]=d[u]+mp[u][i];
if(!inq[i])
q.push(i);
inq[i]=1;
}
}
}
return d[t];
}

int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i = 0;i<n;i++)
for(int j = 0;j<n;j++)
scanf("%d",&mp[i][j]);
int res1 = inf;
int ans = spfa(0,n-1,res1);
int res2 = inf;
spfa(n-1,0,res2);
printf("%d\n",min(ans,res1+res2));
}
}


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<i<n), satisfies ∑X ki (1<=k<=n)=∑X ij (1<=j<=n). 

For example, if n=4,we can get the following equality: 

X 12+X 13+X 14=1 

X 14+X 24+X 34=1 

X 12+X 22+X 32+X 42=X 21+X 22+X 23+X 24 

X 13+X 23+X 33+X 43=X 31+X 32+X 33+X 34 

Now ,we want to know the minimum of ∑C ij*X ij(1<=i,j<=n) you can get. 

Hint

For sample, X 12=X 24=1,all other X ij is 0. 

Input

The input consists of multiple test cases (less than 35 case). 

For each test case ,the first line contains one integer n (1<n<=300). 

The next n lines, for each lines, each of which contains n integers, illustrating the matrix C, The j-th integer on i-th line is C ij(0<=Cij<=100000).

Output

For each case, output the minimum of ∑C ij*X ij you can get. 

Sample Input

4
1 2 4 10
2 0 1 1
2 2 0 5
6 3 1 2

Sample Output

3