bzoj1641[Usaco2007 Nov]Cow Hurdles 奶牛跨栏
题意：
n点m边有向图，每次给出询问x，y求x到y路径中最大边权的最小值是多少。n≤500
题解：
floyd变形。
代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #define inc(i,j,k) for(int i=j;i<=k;i++) 5 #define maxn 310 6 #define INF 0x3fffffff 7 using namespace std; 8 9 inline int read(){ 10 char ch=getchar(); int f=1,x=0; 11 while(ch<'0'||ch>'9'){if(ch=='-')f=-1; ch=getchar();} 12 while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar(); 13 return f*x; 14 } 15 int n,m,t,g[maxn][maxn]; 16 int main(){ 17 n=read(); m=read(); t=read(); inc(i,1,n)inc(j,1,n)g[i][j]=INF; 18 inc(i,1,m){int a=read(),b=read(); g[a][b]=read();} 19 inc(k,1,n)inc(i,1,n)inc(j,1,n)g[i][j]=min(g[i][j],max(g[i][k],g[k][j])); 20 inc(i,1,t){int a=read(),b=read(); printf("%d ",g[a][b]==INF?-1:g[a][b]);} 21 return 0; 22 }
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