floyd骚操作——传递闭包

Posted dillonh

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传递闭包的含义指通过传递性推导出尽量多的元素之间的关系,而传递闭包一般都是采用floyd算法。

下面用两道题来实现传递闭包:

Problem 1(POJ3660):

题目链接:http://poj.org/problem?id=3660

题目:

技术分享图片

技术分享图片

题意:n头牛参加比赛,给你m对关系(譬如给你a和b,那么给的就是a必赢b,当然,b能赢c,那么a也能赢c),问能确定多少头牛的排名。

思路:首先我们用flod算法将所有的关系进行传递,只要u能胜v,那么我们就将d[u][v]设为1,最后如果两者之间有d[u][v]=1或d[v][u]且二者不能同时出现时ans++。

代码实现如下:

技术分享图片
 1 #include <set>
 2 #include <map>
 3 #include <queue>
 4 #include <stack>
 5 #include <cmath>
 6 #include <bitset>
 7 #include <cstdio>
 8 #include <string>
 9 #include <vector>
10 #include <cstdlib>
11 #include <cstring>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 
16 typedef long long ll;
17 typedef pair<ll, ll> pll;
18 typedef pair<int, ll> pil;;
19 typedef pair<int, int> pii;
20 typedef unsigned long long ull;
21 
22 #define lson i<<1
23 #define rson i<<1|1
24 #define bug printf("*********
");
25 #define FIN freopen("D://code//in.txt", "r", stdin);
26 #define debug(x) cout<<"["<<x<<"]" <<endl;
27 #define IO ios::sync_with_stdio(false),cin.tie(0);
28 
29 const double eps = 1e-8;
30 const int mod = 10007;
31 const int maxn = 4500 + 7;
32 const double pi = acos(-1);
33 const int inf = 0x3f3f3f3f;
34 const ll INF = 0x3f3f3f3f3f3f3f;
35 
36 int n, m, u, v;
37 int relationship[107][107];
38 
39 int main() {
40     //FIN;
41     scanf("%d%d", &n, &m);
42     memset(relationship, 0, sizeof(relationship));
43     for(int i = 1; i <= m; i++) {
44         scanf("%d%d", &u, &v);
45         relationship[u][v] = 1;
46     }
47     for(int k = 1; k <= n; k++) {
48         for(int i = 1; i <= n; i++) {
49             for(int j = 1; j <= n; j++) {
50                 if(relationship[i][k] && relationship[k][j]) {
51                     relationship[i][j] = 1;
52                 }
53             }
54         }
55     }
56     int ans = 0, j;
57     for(int i = 1; i <= n; i++) {
58         for(j = 1; j <= n; j++) {
59             if(i == j) continue;
60             if(relationship[i][j] == 0 && relationship[j][i] == 0) {
61                 break;
62             }
63         }
64         if(j > n) ans++;
65     }
66     printf("%d
", ans);
67     return 0;
68 }
View Code

 

Problem 2(POJ1094)

题目链接:http://poj.org/problem?id=1094

题目:

技术分享图片

技术分享图片

题意:给你n个大写字母,m对大小关系,根据他给的关系推测是否有大小矛盾的情况。如果有矛盾的就输出是在第几组关系时矛盾;如果不矛盾,判断只需要前t对组关系就能推测出他们从小到大的排序;如果没有以上两种情况就输入无法确定。

思路:对于每输入一对关系就跑一次floyd判断一遍,如果能推测出他们的关系,那么就跑一边拓扑排序求出他们从小打到的排序情况;如果有矛盾的关系就直接输出是在第几组关系时矛盾;如果没有以上情况就输出无法确定。

代码实现如下:

技术分享图片
  1 #include <set>
  2 #include <map>
  3 #include <queue>
  4 #include <stack>
  5 #include <cmath>
  6 #include <bitset>
  7 #include <cstdio>
  8 #include <string>
  9 #include <vector>
 10 #include <cstdlib>
 11 #include <cstring>
 12 #include <iostream>
 13 #include <algorithm>
 14 using namespace std;
 15 
 16 typedef long long ll;
 17 typedef pair<ll, ll> pll;
 18 typedef pair<int, ll> pil;;
 19 typedef pair<int, int> pii;
 20 typedef unsigned long long ull;
 21 
 22 #define lson i<<1
 23 #define rson i<<1|1
 24 #define bug printf("*********
");
 25 #define FIN freopen("D://code//in.txt", "r", stdin);
 26 #define debug(x) cout<<"["<<x<<"]" <<endl;
 27 #define IO ios::sync_with_stdio(false),cin.tie(0);
 28 
 29 const double eps = 1e-8;
 30 const int mod = 10007;
 31 const int maxn = 4500 + 7;
 32 const double pi = acos(-1);
 33 const int inf = 0x3f3f3f3f;
 34 const ll INF = 0x3f3f3f3f3f3f3f;
 35 
 36 int n, m, t, flag;
 37 char s[1007][5];
 38 int d[30][30], in[30], num[30];
 39 vector<int> G[30];
 40 
 41 bool floyd() {
 42     for(int k = 1; k <= n; k++) {
 43         for(int i = 1; i <= n; i++) {
 44             for(int j = 1; j <= n; j++) {
 45                 if(d[i][k] && d[k][j]) {
 46                     d[i][j] = 1;
 47                 }
 48             }
 49         }
 50     }
 51     for(int i = 1; i <= n; i++) {
 52         for(int j = 1; j <= n; j++) {
 53             if(i == j) continue;
 54             if((d[i][j] && d[j][i]) || (d[i][j] == 0 && d[j][i] == 0)) {
 55                 return false;
 56             }
 57         }
 58     }
 59     return true;
 60 }
 61 
 62 void topsort(int m) {
 63     t = 0;
 64     for(int i = 1; i <= n; i++) {
 65         G[i].clear();
 66     }
 67     memset(in, 0, sizeof(in));
 68     for(int i = 1; i <= m; i++) {
 69         int x = s[i][0] - A + 1, y = s[i][2] - A + 1;
 70         G[x].push_back(y);
 71         in[y]++;
 72     }
 73     queue<int> q;
 74     for(int i = 1; i <= n; i++) {
 75         if(in[i] == 0) {
 76             q.push(i);
 77         }
 78     }
 79     while(!q.empty()) {
 80         int x = q.front(); q.pop();
 81         num[t++] = x;
 82         for(int i = 0; i < G[x].size(); i++) {
 83             int v = G[x][i];
 84             in[v]--;
 85             if(in[v] == 0) {
 86                 q.push(v);
 87             }
 88         }
 89     }
 90 }
 91 
 92 int main() {
 93     //FIN;
 94     while(~scanf("%d%d", &n, &m)) {
 95         if(n == 0 && m == 0) break;
 96         memset(d, 0, sizeof(d));
 97         for(int i = 1; i <= m; i++) {
 98             scanf("%s", s[i]);
 99         }
100         flag = 0;
101         for(int i = 1; i <= m; i++) {
102             int x = s[i][0] - A + 1, y = s[i][2] - A + 1;
103             d[x][y] = 1;
104             if(floyd()) {
105                 printf("Sorted sequence determined after %d relations: ", i);
106                 topsort(i);
107                 for(int i = 0; i < t; i++) {
108                     printf("%c", num[i] - 1 + A);
109                 }
110                 printf(".
");
111                 flag = 1;
112             } else {
113                 for(int j = 1; j <= n; j++) {
114                     for(int k = 1; k <= n; k++) {
115                         if(j == k) continue;
116                         if((d[j][k] && d[k][j])) {
117                             printf("Inconsistency found after %d relations.
", i);
118                             flag = 1;
119                             break;
120                         }
121                     }
122                     if(flag) break;
123                 }
124             }
125             if(flag) break;
126         }
127         if(!flag) printf("Sorted sequence cannot be determined.
");
128     }
129     return 0;
130 }
View Code

 

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