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 << "]" << "\n";
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 * (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 << "]" << "\n";
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 * (N - 1) / 2 - cnt);
139 return 0;
140 }
原文地址：https://www.cnblogs.com/zaq19970105/p/11005295.html
时间： 2024-10-31 02:11:25