HDU 4370 0 or 1（最短路）

转载自：http://blog.csdn.net/qq_21057881/article/details/52128980

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

41 2 4 102 0 1 12 2 0 56 3 1 2

Sample Output

3

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 <iostream>  #include<cstdio>  #include<queue>  #include<cstring>  #include<algorithm>  using namespace std;  #define inf 1e9const 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));}}