Rank of Tetris HDU1811 并查集+拓扑排序

 

题意:有A>B A=B A<B 三种情况,是否能通过多次判断出各自排名

Sample Input

3 3
0 > 1
1 < 2
0 > 2
4 4
1 = 2
1 > 3
2 > 0
0 > 1
3 3
1 > 0
1 > 2
2 < 1

Sample Output

OK
CONFLICT
UNCERTAIN
#include<iostream>
#include<algorithm>
#include<string>
#include<cmath>
#include<cstring>
#include<set>
#include<map>
#include<queue>
#include<time.h>
#include<cstdio>
#include<stack>
using namespace std; 
typedef long long ll;
const int maxn = 1000000;
const int mod = 1e9+7;
#define pb push_back
// #define long long int
#define INF 0x3f3f3f3f;
using namespace std;
/*
    2020.10.15
*/
int pre[maxn];
int mk[maxn];
int n,m;
int sum;
int a[maxn];
int b[maxn];
char op[maxn];
vector<int> nex[maxn];

int find(int x)
{
    if(x!=pre[x])
        pre[x] = find(pre[x]);
    return pre[x];
}
void init()
{
    for(int i=0;i<n;i++)
    {
        pre[i] = i;
        mk[i] = 0;
        nex[i].clear();
    }
}
bool uoin(int x,int y)
{
    int u = find(x);
    int v = find(y);
    if(u==v)
        return true;

    pre[u] = v;
    return false;
}
void topsort()
{
    queue<int> q;
    bool flag = 0;


    for(int i=0;i<n;i++)
        if( mk[i] == 0 && pre[i] == i)
            q.push(i);
        
    
    while(!q.empty())
    {
        if(q.size()>1) flag = 1;
        int cur = q.front();

        q.pop();
        sum--;

        for(int i=0;i<nex[cur].size();i++)
        {
            if(--mk[nex[cur][i]]==0)
            {
                q.push(nex[cur][i]);
            }
        }
    }
    if( sum>0 ) printf("CONFLICT\n");
    else if(flag) printf("UNCERTAIN\n");
    else printf("OK\n");
}
int main()
{
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        init();
        sum = n;
        for(int i=0;i<m;i++)
        {
            scanf("%d %c %d",&a[i],&op[i],&b[i]);
            if(op[i]=='=')
            {
                if(!uoin(a[i],b[i]))
                {
                    sum--;
                }
            }
        }

        for(int i=0;i<m;i++)
        {
            if(op[i]=='=')continue;
        
            int u = find(a[i]);
            int v = find(b[i]);

            if(op[i]=='>')
            {
                nex[u].pb(v);
                mk[v]++;
            }
            else
            {
                nex[v].pb(u);
                mk[u]++;
            }
        }
        topsort();
    }
    //system("pause");
}
View Code

题解: A=B时候 并查集uoin 拓扑排序总数 - 1 , A>B A<B 情况进行拓扑排序

mk[i] 表示该点入度

 

posted @ 2020-10-17 13:11  Wh1te  阅读(91)  评论(0编辑  收藏  举报