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Description
Input
Output
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
一开始瞎做过不了 然后看了别人的代码才发现要用并查集 把=de的两个化为一个就好
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w",stdout);
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long LL;
const int MAXN=1e5+5;
struct Edge{
int v,nxt;
}E[MAXN*2];
int Head[MAXN],erear;
int IN[MAXN],P[MAXN],sz;
int Same[MAXN];
int A[MAXN],B[MAXN];
char OP[MAXN];
int Father[MAXN],Rank[MAXN];
int n,m;
void all_init(){
erear=0;
memset(Head,-1,sizeof(Head));
memset(IN,0,sizeof(IN));
for(int i=0;i<n;i++){
Father[i]=i;
}
}
int Find(int x){
return Father[x]==x?x:(Father[x]=Find(Father[x]));
}
void edge_add(int u,int v){
E[erear].v=v;
E[erear].nxt=Head[u];
Head[u]=erear++;
}
int main()
{
//FIN
char op;
while(~scanf("%d%d",&n,&m))
{
int sum=n;
all_init();
for(int i=0;i<m;i++){
scanf("%d %c %d",&A[i],&OP[i],&B[i]);
if(OP[i]==‘=‘){
int root1=Find(A[i]);
int root2=Find(B[i]);
if(root1!=root2){
Father[root2]=root1;
sum--;
}
}
}
for(int i=0;i<m;i++){
if(OP[i]==‘=‘){
continue;
}
else if(OP[i]==‘>‘){
int u=Find(A[i]);
int v=Find(B[i]);
edge_add(u,v);
IN[v]++;
}
else if(OP[i]==‘<‘){
int u=Find(A[i]);
int v=Find(B[i]);
edge_add(v,u);
IN[u]++;
}
}
sz=0;
queue<int>Q;
while (!Q.empty()) Q.pop();
for(int i=0;i<n;i++){
if(!IN[i]&&Find(i)==i) Q.push(i);
}
bool sign=0;
while(!Q.empty()){
if(Q.size()>=2) sign=1;
int u=Q.front();
Q.pop();
P[sz++]=u;
sum--;
for(int i=Head[u];i!=-1;i=E[i].nxt){
int v=E[i].v;
IN[v]--;
if(!IN[v]) Q.push(v);
}
}
/*for(int i=0;i<sz;i++){
if(i==0) printf("%d",P[i]);
else printf(" %d",P[i]);
}
printf("\n");
cout<<"-----------------sz="<<sz<<endl;
cout<<sum<<endl;*/
if(sum>0){
printf("CONFLICT\n");
}
else if(sign){
printf("UNCERTAIN\n");
}
else{
printf("OK\n");
}
}
return 0;
}
HDU 1811 Rank of Tetris 拓扑排序+并查集
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原文地址:http://www.cnblogs.com/Hyouka/p/5743167.html