HDU 4370 0 or 1

0 or 1

Time Limit: 2000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 4370
64-bit integer IO format: %I64d      Java class name: Main

Given a n*n matrix Cij (1<=i,j<=n),We want to find a n*n matrix $X_{ij} (1<=i,j<=n)$,which is 0 or 1.

Besides,Xij 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 [sum{X_{ki}} (1<=k<=n)=sum{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 $sum{C_{ij} imes X_{ij}}(1<=i,j<=n)$ 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 $sum{C_{ij} imes X_{ij}}$ 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

Source

2012 Multi-University Training Contest 8

 

解题：居然可以转化为最短路 惊呆了。。

 

观察等式我们可以发现 1的出度为1 n的入度为1 其余的入度等于出度

 

01矩阵 2-n-1行 行和等于列和

 

可以求从1到n的最短路，这个还可以理解，那个1到其他再到1 n到其他再到n 暂时我没法理解

 

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
const int maxn = 310;
int n,mp[maxn][maxn],d[maxn];
int spfa(int s,int t,int &val) {
    queue<int>q;
    bool inq[maxn] = {false};
    memset(d,0x3f,sizeof d);
    q.push(s);
    d[s] = 0;
    val = INT_MAX;
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        inq[u] = false;
        for(int i = 0; i < n; ++i) {
            if(u != s && i == s) val = min(val,d[u]+mp[u][i]);
            if(d[i] > d[u] + mp[u][i]) {
                d[i] = d[u] + mp[u][i];
                if(!inq[i]) {
                    inq[i] = true;
                    q.push(i);
                }
            }
        }
    }
    return d[t];
}
int main() {
    while(~scanf("%d",&n)) {
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < n; ++j)
                scanf("%d",mp[i]+j);
        int a,b;
        int ret = spfa(0,n-1,a);
        spfa(n-1,0,b);
        printf("%d
",min(a+b,ret));
    }
    return 0;
}