0 or 1 HDU - 4370(最短路）

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

题意：表示1号点出度为1，表示n号点入度为1，表示k( 1 < k < n )号点出度等于入度。∑Cij*Xij(1<=i,j<=n)，很明显，这是某个路径的花费，而路径的含义可以有以下两种：

一：1号点到n号点的花费

二：1号点经过其它点成环，n号点经过其它点成环，这两个环的花费之和

于是，就变成了一道简单的最短路问题

关于环花费的算法，可以改进spfa算法，初始化dis[start] = INF，且一开始让源点之外的点入队

4

1 2 4 10

2 0 1 1

2 2 0 5

6 3 1 2

Sample Output

3

Input

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

For each test case ,the first line contains one integer n (1

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<stack>
#define INF 0x3f3f3f3f
using namespace std;
int cost[550][550];
int dis[500],vis[500];
int n;
int min(int a,int b)
{
return  a < b ? a : b;
}
void spfa(int start)
{
stack<int>Q;
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)
{
dis[i] = cost[start][i];
if(i!=start)
{
Q.push(i);
vis[i] = 1;
}
}
dis[start] = INF;
while(!Q.empty())
{
int x = Q.top();
Q.pop();
vis[x] = 0;
for(int y = 1;y<=n;y++)
{
if(x==y) continue;
if(dis[x]+cost[x][y]<dis[y])
{
dis[y] = dis[x]+cost[x][y];
if<
4000
/span>(!vis[y])
{
vis[y] = 1;
Q.push(y);
}
}
}
}
}
int main()
{
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
scanf("%d",&cost[i][j]);
}
}
int ans,c1,cn;
spfa(1);
ans = dis
;
c1  = dis[1];//从１到１
spfa(n);
cn = dis
;//从n到n
ans = min(ans,c1+cn);
printf("%d\n",ans);
}
return 0;
}