HDOJ 1811 Rank of Tetris (并查集+topo)

题意

给出一些人之间的rating高低或等于关系，判断给出的关系是否冲突，能不能通过这些排出所有人。

思路

如果没有等于的话这个题是个比较经典的拓扑排序，加上等于之后我们每次相等的两边的关系需要合并，所以用并查集专门维护一下相等时的关系集合，其他的拓扑排序来做即可。

代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
using namespace std;
#define LL long long
#define Lowbit(x) ((x)&(-x))
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1|1
#define MP(a, b) make_pair(a, b)
const int INF = 0x3f3f3f3f;
const int Mod = 1000000007;
const int maxn = 10000 + 7;
const double eps = 1e-8;
const double PI = acos(-1.0);
typedef pair<int, int> pii;

int n, m;
int fa[maxn];
int a[maxn], b[maxn];
char op[maxn];
int in[maxn], cnt, flag;
vector<int> G[maxn];

void init()
{
for (int i = 0; i <= n; i++)
fa[i] = i;
for (int i = 0; i <= n; i++)
in[i] = 0, G[i].clear();
flag = 0;
cnt = n;
}

int find(int x)
{
return fa[x] == x ? x : fa[x] = find(fa[x]);
}

bool merge(int a, int b)
{
int u = find(a);
int v = find(b);
if (u == v) return false;
if (u > v) fa[v] = u;
else fa[u] = v;
return true;
}

void toposort()
{
queue<int> q;
for (int i = n - 1; i >= 0; i--)
if (!in[i] && find(i) == i)
q.push(i);
while (!q.empty())
{
if (q.size() > 1) flag = 1;
int u = q.front(); q.pop();
cnt--;
for (int i = 0; i < G[u].size(); i++)
{
in[G[u][i]]--;
if (in[G[u][i]] == 0)
q.push(G[u][i]);
}
}
}

int main()
{
//freopen("H:\\in.txt","r",stdin);
//freopen("H:\\out.txt","w",stdout);
while (scanf("%d%d", &n, &m) != EOF)
{
init();
for (int i = 0; i < m; i++)
{
scanf("%d %c %d", &a[i], &op[i], &b[i]);
if (op[i] == '=' && merge(a[i], b[i]))
cnt--;
}
for (int i = 0; i < m; i++)
{
int u = find(a[i]);
int v = find(b[i]);
if (op[i] == '<')
{
G[v].push_back(u);
in[u]++;
}
else if (op[i] == '>')
{
G[u].push_back(v);
in[v]++;
}
}
flag = 0;
toposort();
if (cnt > 0) puts("CONFLICT");
else if (flag) puts("UNCERTAIN");
else puts("OK");
}
return 0;
}