USACO 2007 “March Gold” Ranking the Cows

题目链接：https://www.luogu.org/problemnew/show/P2881

题目链接：https://vjudge.net/problem/POJ-3275

题目大意

　　给定标号为 1~N 这 N 个数，在给定 M 组大小关系，求还需要知道多少组大小关系才可以给这组数排序？

分析1(Floyd + bitset)

　　总共需要知道 n * (n - 1) / 2 条边，因此只要求一下现在已经有了多少条边，再减一下即可。由于大小关系有传递性，因此计数之前需要求传递闭包。

　　直接上 floyd($O(n^3)​$) 会超时，需要用bitset或手动压位，可以在$O(\frac{n^3}{w})​$求出传递闭包，其中w表示字长，为64或32。

代码如下

 #include <bits/stdc++.h>
 using namespace std;

 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 #define Rep(i,n) for (int i = 0; i < (n); ++i)
 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

 #define pr(x) cout << #x << " = " << x << "  "
 #define prln(x) cout << #x << " = " << x << endl

 #define LOWBIT(x) ((x)&(-x))

 #define ALL(x) x.begin(),x.end()
 #define INS(x) inserter(x,x.begin())

 #define ms0(a) memset(a,0,sizeof(a))
 #define msI(a) memset(a,inf,sizeof(a))
 #define msM(a) memset(a,-1,sizeof(a))

 #define MP make_pair
 #define PB push_back
 #define ft first
 #define sd second

 template<typename T1, typename T2>
 istream &operator>>(istream &in, pair<T1, T2> &p) {
     in >> p.first >> p.second;
     return in;
 }

 template<typename T>
 istream &operator>>(istream &in, vector<T> &v) {
     for (auto &x: v)
         in >> x;
     return in;
 }

 template<typename T1, typename T2>
 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
     out << "[" << p.first << ", " << p.second << "]" << "\n";
     return out;
 }

 inline int gc(){
     static const int BUF = 1e7;
     static char buf[BUF], *bg = buf + BUF, *ed = bg;

     if(bg == ed) fread(bg = buf, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , c=gc());
     return x*f;
 }

 typedef long long LL;
 typedef unsigned long long uLL;
 typedef pair< double, double > PDD;
 typedef pair< int, int > PII;
 typedef pair< int, PII > PIPII;
 typedef pair< string, int > PSI;
 typedef pair< int, PSI > PIPSI;
 typedef set< int > SI;
 typedef vector< int > VI;
 typedef vector< VI > VVI;
 typedef vector< PII > VPII;
 typedef map< int, int > MII;
 typedef map< int, PII > MIPII;
 typedef map< string, int > MSI;
 typedef multimap< int, int > MMII;
 //typedef unordered_map< int, int > uMII;
 typedef pair< LL, LL > PLL;
 typedef vector< LL > VL;
 typedef vector< VL > VVL;
 typedef priority_queue< int > PQIMax;
 typedef priority_queue< int, VI, greater< int > > PQIMin;
 const double EPS = 1e-;
 const LL inf = 0x7fffffff;
 const LL infLL = 0x7fffffffffffffffLL;
 const LL mod = 1e9 + ;
 const int maxN = 1e3 + ;
 const LL ONE = ;
 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 const LL oddBits = 0x5555555555555555;

 int N, M, cnt;
 bitset< maxN > m[maxN];

 int main(){
     //freopen("MyOutput.txt","w",stdout);
     //freopen("input.txt","r",stdin);
     INIT();
     N = ri();
     M = ri();
     For(i, , N) m[i][i] = ;
     Rep(i, M) {
         int x, y;
         x = ri();
         y = ri();
         m[x][y] = ;
     }

     For(i, , N) For(j, , N) if(m[j][i]) m[j] |= m[i];
     For(i, , N) cnt += m[i].count();
     cnt -= N; // 减去 i->i 的，有 N 条 

     printf("%d\n", N * (N - ) /  - cnt);
     return ;
 }

分析2(dfs+ bitset)

　　考虑到通过压位过的邻接矩阵求传递闭包时做了很多多余的操作，我们可以用邻接链表来求传递闭包，然后用邻接矩阵计数。复杂度可以降到$O(\frac{n * (n + m)}{w})​$。

代码如下

 #include <bits/stdc++.h>
 using namespace std;

 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 #define Rep(i,n) for (int i = 0; i < (n); ++i)
 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

 #define pr(x) cout << #x << " = " << x << "  "
 #define prln(x) cout << #x << " = " << x << endl

 #define LOWBIT(x) ((x)&(-x))

 #define ALL(x) x.begin(),x.end()
 #define INS(x) inserter(x,x.begin())

 #define ms0(a) memset(a,0,sizeof(a))
 #define msI(a) memset(a,inf,sizeof(a))
 #define msM(a) memset(a,-1,sizeof(a))

 #define MP make_pair
 #define PB push_back
 #define ft first
 #define sd second

 template<typename T1, typename T2>
 istream &operator>>(istream &in, pair<T1, T2> &p) {
     in >> p.first >> p.second;
     return in;
 }

 template<typename T>
 istream &operator>>(istream &in, vector<T> &v) {
     for (auto &x: v)
         in >> x;
     return in;
 }

 template<typename T1, typename T2>
 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
     out << "[" << p.first << ", " << p.second << "]" << "\n";
     return out;
 }

 inline int gc(){
     static const int BUF = 1e7;
     static char buf[BUF], *bg = buf + BUF, *ed = bg;

     if(bg == ed) fread(bg = buf, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , c=gc());
     return x*f;
 }

 typedef long long LL;
 typedef unsigned long long uLL;
 typedef pair< double, double > PDD;
 typedef pair< int, int > PII;
 typedef pair< int, PII > PIPII;
 typedef pair< string, int > PSI;
 typedef pair< int, PSI > PIPSI;
 typedef set< int > SI;
 typedef vector< int > VI;
 typedef vector< VI > VVI;
 typedef vector< PII > VPII;
 typedef map< int, int > MII;
 typedef map< int, PII > MIPII;
 typedef map< string, int > MSI;
 typedef multimap< int, int > MMII;
 //typedef unordered_map< int, int > uMII;
 typedef pair< LL, LL > PLL;
 typedef vector< LL > VL;
 typedef vector< VL > VVL;
 typedef priority_queue< int > PQIMax;
 typedef priority_queue< int, VI, greater< int > > PQIMin;
 const double EPS = 1e-;
 const LL inf = 0x7fffffff;
 const LL infLL = 0x7fffffffffffffffLL;
 const LL mod = 1e9 + ;
 const int maxN = 1e3 + ;
 const LL ONE = ;
 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 const LL oddBits = 0x5555555555555555;

 struct Edge{
     int from, to;
 };

 struct Vertex{
     VI edges;
 };

 int N, M, cnt;
 bitset< maxN > m[maxN], vis;
 Edge e[maxN << ];
 int elen;
 Vertex v[maxN];

 // 找到 x 号节点所能到达的所有节点
 void dfs(int x) {
     vis[x] = ;
     foreach(i, v[x].edges) { // 结果取决于 x 的孩子节点所能到达的节点，此处相当于 m[x][y] = 1
         int y = e[*i].to;
         if(!vis[y]) dfs(y);
         m[x] |= m[y];
     }
 }

 int main(){
     //freopen("MyOutput.txt","w",stdout);
     //freopen("input.txt","r",stdin);
     INIT();
     N = ri();
     M = ri();
     For(i, , N) m[i][i] = ;
     Rep(i, M) {
         int x, y;
         x = ri();
         y = ri();
         e[++elen].from = x;
         e[elen].to = y;
         v[x].edges.PB(elen);
     }

     For(i, , N) if(!vis[i]) dfs(i);
     For(i, , N) cnt += m[i].count();
     cnt -= N; // 减去 i->i 的，有 N 条 

     printf("%d\n", N * (N - ) /  - cnt);
     return ;
 }