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
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); 5 #define Rep(i,n) for (int i = 0; i < (n); ++i) 6 #define For(i,s,t) for (int i = (s); i <= (t); ++i) 7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) 8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) 9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i) 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 12 13 #define pr(x) cout << #x << " = " << x << " " 14 #define prln(x) cout << #x << " = " << x << endl 15 16 #define LOWBIT(x) ((x)&(-x)) 17 18 #define ALL(x) x.begin(),x.end() 19 #define INS(x) inserter(x,x.begin()) 20 21 #define ms0(a) memset(a,0,sizeof(a)) 22 #define msI(a) memset(a,inf,sizeof(a)) 23 #define msM(a) memset(a,-1,sizeof(a)) 24 25 #define MP make_pair 26 #define PB push_back 27 #define ft first 28 #define sd second 29 30 template<typename T1, typename T2> 31 istream &operator>>(istream &in, pair<T1, T2> &p) { 32 in >> p.first >> p.second; 33 return in; 34 } 35 36 template<typename T> 37 istream &operator>>(istream &in, vector<T> &v) { 38 for (auto &x: v) 39 in >> x; 40 return in; 41 } 42 43 template<typename T1, typename T2> 44 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) { 45 out << "[" << p.first << ", " << p.second << "]" << " "; 46 return out; 47 } 48 49 inline int gc(){ 50 static const int BUF = 1e7; 51 static char buf[BUF], *bg = buf + BUF, *ed = bg; 52 53 if(bg == ed) fread(bg = buf, 1, BUF, stdin); 54 return *bg++; 55 } 56 57 inline int ri(){ 58 int x = 0, f = 1, c = gc(); 59 for(; c<48||c>57; f = c=='-'?-1:f, c=gc()); 60 for(; c>47&&c<58; x = x*10 + c - 48, c=gc()); 61 return x*f; 62 } 63 64 typedef long long LL; 65 typedef unsigned long long uLL; 66 typedef pair< double, double > PDD; 67 typedef pair< int, int > PII; 68 typedef pair< int, PII > PIPII; 69 typedef pair< string, int > PSI; 70 typedef pair< int, PSI > PIPSI; 71 typedef set< int > SI; 72 typedef vector< int > VI; 73 typedef vector< VI > VVI; 74 typedef vector< PII > VPII; 75 typedef map< int, int > MII; 76 typedef map< int, PII > MIPII; 77 typedef map< string, int > MSI; 78 typedef multimap< int, int > MMII; 79 //typedef unordered_map< int, int > uMII; 80 typedef pair< LL, LL > PLL; 81 typedef vector< LL > VL; 82 typedef vector< VL > VVL; 83 typedef priority_queue< int > PQIMax; 84 typedef priority_queue< int, VI, greater< int > > PQIMin; 85 const double EPS = 1e-10; 86 const LL inf = 0x7fffffff; 87 const LL infLL = 0x7fffffffffffffffLL; 88 const LL mod = 1e9 + 7; 89 const int maxN = 1e3 + 7; 90 const LL ONE = 1; 91 const LL evenBits = 0xaaaaaaaaaaaaaaaa; 92 const LL oddBits = 0x5555555555555555; 93 94 int N, M, cnt; 95 bitset< maxN > m[maxN]; 96 97 int main(){ 98 //freopen("MyOutput.txt","w",stdout); 99 //freopen("input.txt","r",stdin); 100 INIT(); 101 N = ri(); 102 M = ri(); 103 For(i, 1, N) m[i][i] = 1; 104 Rep(i, M) { 105 int x, y; 106 x = ri(); 107 y = ri(); 108 m[x][y] = 1; 109 } 110 111 For(i, 1, N) For(j, 1, N) if(m[j][i]) m[j] |= m[i]; 112 For(i, 1, N) cnt += m[i].count(); 113 cnt -= N; // 减去 i->i 的，有 N 条 114 115 printf("%d ", N * (N - 1) / 2 - cnt); 116 return 0; 117 }
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
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); 5 #define Rep(i,n) for (int i = 0; i < (n); ++i) 6 #define For(i,s,t) for (int i = (s); i <= (t); ++i) 7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) 8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) 9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i) 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 12 13 #define pr(x) cout << #x << " = " << x << " " 14 #define prln(x) cout << #x << " = " << x << endl 15 16 #define LOWBIT(x) ((x)&(-x)) 17 18 #define ALL(x) x.begin(),x.end() 19 #define INS(x) inserter(x,x.begin()) 20 21 #define ms0(a) memset(a,0,sizeof(a)) 22 #define msI(a) memset(a,inf,sizeof(a)) 23 #define msM(a) memset(a,-1,sizeof(a)) 24 25 #define MP make_pair 26 #define PB push_back 27 #define ft first 28 #define sd second 29 30 template<typename T1, typename T2> 31 istream &operator>>(istream &in, pair<T1, T2> &p) { 32 in >> p.first >> p.second; 33 return in; 34 } 35 36 template<typename T> 37 istream &operator>>(istream &in, vector<T> &v) { 38 for (auto &x: v) 39 in >> x; 40 return in; 41 } 42 43 template<typename T1, typename T2> 44 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) { 45 out << "[" << p.first << ", " << p.second << "]" << " "; 46 return out; 47 } 48 49 inline int gc(){ 50 static const int BUF = 1e7; 51 static char buf[BUF], *bg = buf + BUF, *ed = bg; 52 53 if(bg == ed) fread(bg = buf, 1, BUF, stdin); 54 return *bg++; 55 } 56 57 inline int ri(){ 58 int x = 0, f = 1, c = gc(); 59 for(; c<48||c>57; f = c=='-'?-1:f, c=gc()); 60 for(; c>47&&c<58; x = x*10 + c - 48, c=gc()); 61 return x*f; 62 } 63 64 typedef long long LL; 65 typedef unsigned long long uLL; 66 typedef pair< double, double > PDD; 67 typedef pair< int, int > PII; 68 typedef pair< int, PII > PIPII; 69 typedef pair< string, int > PSI; 70 typedef pair< int, PSI > PIPSI; 71 typedef set< int > SI; 72 typedef vector< int > VI; 73 typedef vector< VI > VVI; 74 typedef vector< PII > VPII; 75 typedef map< int, int > MII; 76 typedef map< int, PII > MIPII; 77 typedef map< string, int > MSI; 78 typedef multimap< int, int > MMII; 79 //typedef unordered_map< int, int > uMII; 80 typedef pair< LL, LL > PLL; 81 typedef vector< LL > VL; 82 typedef vector< VL > VVL; 83 typedef priority_queue< int > PQIMax; 84 typedef priority_queue< int, VI, greater< int > > PQIMin; 85 const double EPS = 1e-10; 86 const LL inf = 0x7fffffff; 87 const LL infLL = 0x7fffffffffffffffLL; 88 const LL mod = 1e9 + 7; 89 const int maxN = 1e3 + 7; 90 const LL ONE = 1; 91 const LL evenBits = 0xaaaaaaaaaaaaaaaa; 92 const LL oddBits = 0x5555555555555555; 93 94 struct Edge{ 95 int from, to; 96 }; 97 98 struct Vertex{ 99 VI edges; 100 }; 101 102 int N, M, cnt; 103 bitset< maxN > m[maxN], vis; 104 Edge e[maxN << 4]; 105 int elen; 106 Vertex v[maxN]; 107 108 // 找到 x 号节点所能到达的所有节点 109 void dfs(int x) { 110 vis[x] = 1; 111 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] = 1 112 int y = e[*i].to; 113 if(!vis[y]) dfs(y); 114 m[x] |= m[y]; 115 } 116 } 117 118 int main(){ 119 //freopen("MyOutput.txt","w",stdout); 120 //freopen("input.txt","r",stdin); 121 INIT(); 122 N = ri(); 123 M = ri(); 124 For(i, 1, N) m[i][i] = 1; 125 Rep(i, M) { 126 int x, y; 127 x = ri(); 128 y = ri(); 129 e[++elen].from = x; 130 e[elen].to = y; 131 v[x].edges.PB(elen); 132 } 133 134 For(i, 1, N) if(!vis[i]) dfs(i); 135 For(i, 1, N) cnt += m[i].count(); 136 cnt -= N; // 减去 i->i 的，有 N 条 137 138 printf("%d ", N * (N - 1) / 2 - cnt); 139 return 0; 140 }