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
一开始没思路,看了题解还是是不懂.....
首先分析一波题..
我们将这个矩阵看作一个链接矩阵;
所以对应条件就变成
1:起点出度为一
2.终点入度为一
3.除了起点终点外其余点的出度等于入度
0为不选这条边,1为选择这条边,所以 ∑C ij*X ij变为从1到n的最短路 或 从1 n开始俩个自环的和 的最小值
所以..跑spfa就可以了
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
const int maxn=1e4+5;
using namespace std;
int nmap[305][305];
int d[305],vis[305],n;
void spfa(int st)
{
queue<int>q;
mem(vis,0);
for(int i=1;i<=n;i++)
{
d[i]=nmap[st][i];
if(i!=st)
{
q.push(i);
vis[i]=1;
}
else
{
vis[i]=0;
}
}
d[st]=inf;
while(!q.empty())
{
int now=q.front();
q.pop();
vis[now]=0;
for(int i=1;i<=n;i++)
{
int u=now,v=i,w=nmap[now][i];
if(u==i) continue;
if(d[v]>d[u]+w)
{
d[v]=d[u]+w;
if(!vis[v])
{
vis[v]=1;
q.push(v);
}
}
}
}
}
int main(){
while(~scanf("%d",&n))
{
mem(nmap,0);
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
scanf("%d",&nmap[i][j]);
spfa(1);
int ans=d[1],minn=d[n];
spfa(n);
ans+=d[n];
printf("%d
",min(minn,ans));
}
return 0;
}