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<=C_ij<=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