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\n");
128 else if(ret) printf("UNCERTAIN\n");
129 else printf("OK\n");
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 }
时间： 2024-10-27 17:31:33