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 ;

 }

巴特西

USACO 2007 “March Gold” Ranking the Cows

题目大意

分析1(Floyd + bitset)

代码如下

分析2(dfs+ bitset)

代码如下

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