HDU-4370 0 or 1( 最短路 )

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4370

Problem 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.X₁₂+X₁₃+...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:

X₁₂+X₁₃+X₁₄=1
X₁₄+X₂₄+X₃₄=1
X₁₂+X₂₂+X₃₂+X₄₂=X₂₁+X₂₂+X₂₃+X₂₄
X₁₃+X₂₃+X₃₃+X₄₃=X₃₁+X₃₂+X₃₃+X₃₄

Now ,we want to know the minimum of ∑C_ij*X_ij(1<=i,j<=n) you can get.

Hint

For sample, X₁₂=X₂₄=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

 

转换思维的一道题，虽然是在最短路的专题里看见的，但是一开始真想不到是最短路，还以为是一个求min( c[1][n], c[1][k] + c[k][n] )( 1 < k < n )的水题，不过认真看完题解发现转换思维之后确实是一道水题，难就难在转换思维

理解条件之前先转换一下思维，将矩阵C看做描述N个点花费的邻接矩阵

再来看三个条件：

条件一：表示1号点出度为1

条件二：表示n号点入度为1

条件三：表示k( 1 < k < n )号点出度等于入度

最后再来看看题目要求，∑C_ij*X_ij(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

代码如下

 

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
#include<cstdio>
#include<queue>
#include<stack>

using namespace std;

const int INF = 0x3f3f3f3f;

int cost[305][305];
int dis[305];
int n;
bool vis[305];

void spfa( int start ){
    stack<int> Q;

    for( int i = 1; i <= n; i++ ){
        dis[i] = cost[start][i];
        if( i != start ){
            Q.push( i );
            vis[i] = true;
        }
        else{
            vis[i] = false;
        }
    }
    dis[start] = INF;

    while( !Q.empty() ){
        int x = Q.top(); Q.pop(); vis[x] = false;

        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( !vis[y] ){
                    vis[y] = true;
                    Q.push( y );
                }
            }
        }
    }
}

int main(){
    ios::sync_with_stdio( false );

    while( cin >> n ){
        for( int i = 1; i <= n; i++ ){
            for( int j = 1; j <= n; j++ ){
                cin >> cost[i][j];
            }
        }

        int ans, c1, cn;
        spfa( 1 );
        ans = dis[n];
        c1 = dis[1];
        spfa( n );
        cn = dis[n];


        cout << min( ans, c1 + cn ) << endl;
    }

    return 0;
}

 

将矩阵C看做描述N个点花费的邻接矩阵