HDU-4370 0 or 1( 最短路 )

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

[align=left]Problem Description[/align]
Given a n*n matrix Cij (1<=i,j<=n),We want to find a n*n matrix Xij (1<=i,j<=n),which is 0 or 1.

Besides,Xij meets the following conditions:

1.X12+X13+...X1n=1
2.X1n+X2n+...Xn-1n=1
3.for each i (1<i<n), satisfies ∑Xki (1<=k<=n)=∑Xij (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 ∑Cij*Xij(1<=i,j<=n) you can get.

Hint

For sample, X12=X24=1,all other Xij is 0.

[align=left]Input[/align]
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 Cij(0<=Cij<=100000).

[align=left]Output[/align]
For each case, output the minimum of ∑Cij*Xij you can get.

[align=left]Sample Input[/align]

4

1 2 4 10
2 0 1 1

2 2 0 5
6 3 1 2

[align=left]Sample Output[/align]

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]
, c[1][k] + c[k]
)( 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，∑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
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
;
c1 = dis[1];
spfa( n );
cn = dis
;

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