题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1811
求一堆数据的拓扑序。
处理：x>y就是x到y一条边，x<y就是y到x一条边。关键问题是处理x=y的情况。
假如x=y，就有问题了。假如不处理的话，可能会被当成少处理一个点而使结果编程UNCERTAIN。所以我们考虑用并查集来解决这个问题。
选谁当祖先？题中又给了一个其他的量叫做RP值，这个RP值的规律是序号越大RP值越大。这样我们可以在合并的时候，尽可能地将RP值大的数当成本集合的祖先。
1 /* 2 ━━━━━┒ギリギリ♂ eye！ 3 ┓┏┓┏┓┃キリキリ♂ mind！ 4 ┛┗┛┗┛┃＼○／ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8 ┛┗┛┗┛┃ 9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <fstream> 24 #include <cassert> 25 #include <cstdio> 26 #include <bitset> 27 #include <vector> 28 #include <deque> 29 #include <queue> 30 #include <stack> 31 #include <ctime> 32 #include <set> 33 #include <map> 34 #include <cmath> 35 using namespace std; 36 #define fr first 37 #define sc second 38 #define cl clear 39 #define BUG puts("here!!!") 40 #define W(a) while(a--) 41 #define pb(a) push_back(a) 42 #define Rint(a) scanf("%d", &a) 43 #define Rll(a) scanf("%lld", &a) 44 #define Rs(a) scanf("%s", a) 45 #define Cin(a) cin >> a 46 #define FRead() freopen("in", "r", stdin) 47 #define FWrite() freopen("out", "w", stdout) 48 #define Rep(i, len) for(int i = 0; i < (len); i++) 49 #define For(i, a, len) for(int i = (a); i < (len); i++) 50 #define Cls(a) memset((a), 0, sizeof(a)) 51 #define Clr(a, x) memset((a), (x), sizeof(a)) 52 #define Full(a) memset((a), 0x7f7f, sizeof(a)) 53 #define lp p << 1 54 #define rp p << 1 | 1 55 #define pi 3.14159265359 56 #define RT return 57 #define lowbit(x) x & (-x) 58 #define onenum(x) __builtin_popcount(x) 59 typedef long long LL; 60 typedef long double LD; 61 typedef unsigned long long ULL; 62 typedef pair<int, int> pii; 63 typedef pair<string, int> psi; 64 typedef map<string, int> msi; 65 typedef vector<int> vi; 66 typedef vector<LL> vl; 67 typedef vector<vl> vvl; 68 typedef vector<bool> vb; 69 70 typedef struct Edge { 71 int u, v, next; 72 Edge() {} 73 Edge(int uu, int vv) : u(uu), v(vv) { next = -1; } 74 }Edge; 75 const int maxn = 10010; 76 const int maxm = 20020; 77 int n, m; 78 Edge edge[maxm]; 79 int head[maxn]; 80 int hcnt; 81 int pre[maxn]; 82 int in[maxn]; 83 int num, q[maxn], front, tail; 84 int uu[maxn], vv[maxn]; 85 char cc[maxn][5]; 86 87 void adde(int u, int v) { 88 edge[hcnt] = Edge(u, v); 89 edge[hcnt].next = head[u]; 90 head[u] = hcnt++; 91 in[v]++; 92 } 93 94 int find(int x) { 95 RT x == pre[x] ? x : pre[x] = find(pre[x]); 96 } 97 98 int unite(int x, int y) { 99 int fx = find(x); 100 int fy = find(y); 101 if(fx == fy) RT 0; 102 if(x > y) pre[fy] = fx; 103 else pre[fx] = fy; 104 return 1; 105 106 } 107 108 int topo() { 109 front = tail = 0; 110 Rep(i, n) { 111 if(in[i] == 0 && i == find(i)) { 112 q[tail++] = i; 113 } 114 } 115 int ret = 0; 116 while(front < tail) { 117 if(tail - front > 1) ret = 1; 118 int u = q[front++]; 119 --num; 120 for(int i = head[u]; ~i; i=edge[i].next) { 121 int v = edge[i].v; 122 if(--in[v] == 0) { 123 q[tail++] = v; 124 } 125 } 126 } 127 if(num > 0) printf("CONFLICT "); 128 else if(ret) printf("UNCERTAIN "); 129 else printf("OK "); 130 } 131 132 int main() { 133 // FRead(); 134 int u, v; 135 while(~Rint(n) && ~Rint(m)) { 136 Cls(in); Clr(head, -1); hcnt = 0; 137 num = n; 138 Rep(i, n+5) pre[i] = i; 139 Rep(i, m) { 140 Rint(uu[i]); Rs(cc[i]); Rint(vv[i]); 141 if(cc[i][0] == '=') { 142 if(unite(uu[i], vv[i])) num--; 143 } 144 } 145 Rep(i, m) { 146 if(cc[i][0] != '=') { 147 u = find(uu[i]); v = find(vv[i]); 148 if(cc[i][0] == '>') adde(u, v); 149 if(cc[i][0] == '<') adde(v, u); 150 } 151 } 152 topo(); 153 } 154 RT 0; 155 }