Description
农夫约翰有N(1≤N≤1000)头奶牛，每一头奶牛都有一个确定的独一无二的正整数产奶率．约翰想要让这些奶牛按产奶率从高到低排序． 约翰已经比较了M(1≤M≤10000)对奶牛的产奶率，但他发现，他还需要再做一张关于另外C对奶牛的产奶率比较，才能推断出所有奶牛的产奶率排序．请帮他确定C的最小值．
Input
第1行包含两个用空格分开的整数N和M.接下来M行，每行有两个用空格分开的整数X和Y(1≤X，y≤1000)，表示奶牛X的产奶率高于奶牛Y.
Output
C的最小值．
Sample Input
5 5
2 1
1 5
2 3
1 4
3 4
INPUT DETAILS:
FJ is comparing 5 cows and has already determined that cow 2 > cow
1, cow 1 > cow 5, cow 2 > cow 3, cow 1 > cow 4, and cow 3 > cow 4
(where the '>' notation means "produces milk more quickly").
2 1
1 5
2 3
1 4
3 4
INPUT DETAILS:
FJ is comparing 5 cows and has already determined that cow 2 > cow
1, cow 1 > cow 5, cow 2 > cow 3, cow 1 > cow 4, and cow 3 > cow 4
(where the '>' notation means "produces milk more quickly").
Sample Output
3
答案为排名无序的点对个数……
然后就总点对-排名有序点对数咯。
#include<bitset> #include<cstdio> #include<algorithm> using namespace std; int read_p,read_ca,read_f; inline int read(){ read_p=0;read_ca=getchar();read_f=1; while(read_ca<'0'||read_ca>'9') read_f=read_ca=='-'?-1:read_f,read_ca=getchar(); while(read_ca>='0'&&read_ca<='9') read_p=read_p*10+read_ca-48,read_ca=getchar(); return read_p*read_f; } int n,m,mmh,x,y; bool bo[1000]; bitset <1000> b[1000]; void dfs(int x){ if (bo[x]) return;bo[x]=1; for (int i=0;i<n;i++) if (b[x][i]) dfs(i),b[x]|=b[i]; } int main(){ register int i; n=read();m=read();mmh=(n*(n-1))>>1; for (i=1;i<=m;i++) x=read()-1,y=read()-1,b[x][y]=1; for (i=0;i<n;i++) dfs(i),mmh-=b[i].count(); printf("%d ",mmh); }