題目大意：
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 (1hint
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 (1output
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
3此題非常厲害，就演算法來說不是很難，但很考想法。這題的精髓在於如何將條件抽象出來，並建圖。
三個條件分別告訴我們 點1的出度為1，點n的入度為1，每個點的入度等於出度，這樣就可以構圖了。設n各點，將∑cij*xij(1<=i,j<=n)看成n各點的鄰接矩陣。這時問題分成了兩種情況。
第一種是求1到n的最短路徑，第二種是求1到1的非自環閉環 + n到n的非自環閉環的和。比較這兩種情況，取最小的值為解。
用spfa時設d[s]的值為inf，讓d[s]的相鄰節點入棧，這樣保證了可以取到非自環閉環。
**：
#include #include #include #include #include #include #include #include #include#include #include #include #include #define ri(n) scanf("%d",&n)
#define oi(n) printf("%d\n",n)
#define rl(n) scanf("%lld",&n)
#define ol(n) printf("%lld\n",n)
#define rep(i,l,r) for(i=l;i<=r;i++)
#define rep1(i,l,r) for(i=l;ip;
while(!p.empty())
p.pop();
for(int i=1; i<=n; i++)
else
}//vis[s]=1;
// c[s]++;
//d[s]=0;
while(!p.empty())
}//}}}
//return true;
}int main()
return 0;
}
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