題目鏈接
題意:就是給你一堆關係,看能不能排出個確定的順序
做法:
1. 拓撲排序+並查集
應該很容易想到的一種思路,大於小於建立單向邊。對於相等的呢,就把他們縮成一個點。就用並查集縮成一個點就行了
- 入度爲0進隊列,隊列數目>1,有衝突
- 沒有遍歷完,信息不完整
- 不然,輸出OK
//#pragma comment (linker, "/STACK:102400000,102400000")
#include<bits/stdc++.h>
#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<set>
#include<stack>
#include<vector>
#include<map>
#include<queue>
#include<list>
#include<time.h>
#define myself i,l,r
#define lson i<<1
#define rson i<<1|1
#define Lson i<<1,l,mid
#define Rson i<<1|1,mid+1,r
#define half (l+r)/2
#define lowbit(x) x&(-x)
#define min4(a, b, c, d) min(min(a,b),min(c,d))
#define min3(x, y, z) min(min(x,y),z)
#define max3(x, y, z) max(max(x,y),z)
#define max4(a, b, c, d) max(max(a,b),max(c,d))
//freopen("E://1.in","r",stdin);
//freopen("E://1.out","w",stdout);
typedef unsigned long long ull;
typedef long long ll;
#define pii make_pair
#define pr pair<int,int>
const int inff = 0x3f3f3f3f;
const int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
const int mdir[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, -1, 1, 1, -1, -1, -1};
const double eps = 1e-10;
const double PI = acos(-1.0);
const double E = 2.718281828459;
using namespace std;
const long long inFF = 9223372036854775807;
const int mod=1e9+7;
const int maxn=1e5+5;
struct nod
{
int x,y;
}a[maxn];
vector<int> v[maxn];
int n,m;
int f[maxn],in[maxn];
int find(int x)
{
return x==f[x]?x:f[x]=find(f[x]);
}
void init()
{
for(int i=0;i<=n;i++) f[i]=i;
for(int i=0;i<=n;i++) v[i].clear(),in[i]=0;
}
int main()
{
while(cin>>n>>m)
{
if(m==0) {
if (n == 0 || n == 1) puts("OK");
else puts("UNCERTAIN");
continue;
}
int cnt=0;
int x,y;
char s[2];
init();
for(int i=1;i<=m;i++)
{
scanf("%d%s%d",&x,s,&y);
if(s[0]=='=')
{
x=find(x);
y=find(y);
if(x!=y) f[x]=y;
}
else if(s[0]=='>') a[++cnt]=nod{x,y};
else a[++cnt]=nod{y,x};
}
int fg=0;
for(int i=1;i<=cnt;i++)
{
int x=find(a[i].x);
int y=find(a[i].y);
if(x==y)
{
fg=1;
break;
}
v[x].push_back(y);
in[y]++;
}
queue<int> q;
int num=0,d=0;
for(int i=0;i<n;i++)
{
if(find(i)==i)
{
d++;
if(in[i]==0) q.push(i);
}
}
if(fg) {puts("CONFLICT");continue;}
while(!q.empty())
{
if(q.size()>1) fg=1;//信息不足
int u=q.front();
q.pop();
num++;
for(int x:v[u])
{
in[x]--;
if(in[x]==0)
q.push(x);
}
}
if(num!=d) puts("CONFLICT");
else if(fg==1) puts("UNCERTAIN");
else puts("OK");
}
}
2.差分約束最短路+並查集
因爲上邊所述的關係屬於差分約束條件,考慮用差分約束來搞。
怎麼構造差分直接跳——>點我,用並查集來判斷是否有沒有信息不完整
造完圖後就是一個最短路判環
如果有負環,衝突
如果
//#pragma comment (linker, "/STACK:102400000,102400000")
#include<bits/stdc++.h>
#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<set>
#include<stack>
#include<vector>
#include<map>
#include<queue>
#include<list>
#include<time.h>
#define myself i,l,r
#define lson i<<1
#define rson i<<1|1
#define Lson i<<1,l,mid
#define Rson i<<1|1,mid+1,r
#define half (l+r)/2
#define lowbit(x) x&(-x)
#define min4(a, b, c, d) min(min(a,b),min(c,d))
#define min3(x, y, z) min(min(x,y),z)
#define max3(x, y, z) max(max(x,y),z)
#define max4(a, b, c, d) max(max(a,b),max(c,d))
//freopen("E://1.in","r",stdin);
//freopen("E://1.out","w",stdout);
typedef unsigned long long ull;
typedef long long ll;
#define pii make_pair
#define pr pair<int,int>
const int inff = 0x3f3f3f3f;
const int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
const int mdir[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, -1, 1, 1, -1, -1, -1};
const double eps = 1e-10;
const double PI = acos(-1.0);
const double E = 2.718281828459;
using namespace std;
const long long inFF = 9223372036854775807;
const int mod=1e9+7;
const int maxn=1e5+5;
int d[maxn],vis[maxn],f[maxn],cnt[maxn];
int head[maxn],sign;
struct node
{
int to,p,val;
}edge[maxn];
int n,m,fg;
void init()
{
for(int i=0;i<=n;i++)
{
f[i]=i;
head[i]=-1;
cnt[i]=0;
}
sign=0;
}
void add(int u,int v,int val)
{
edge[sign]=node{v,head[u],val};
head[u]=sign++;
}
int find(int x)
{
return x==f[x]?x:f[x]=find(f[x]);
}
void meger(int x,int y)
{
int fx=find(x);
int fy=find(y);
if(fx!=fy)
f[x]=y;//x->y 有一點疑惑,不知道爲什麼這麼合併,但是確實能判斷信息是否完整
}
void spfa()
{
for(int i=0;i<=n;i++) d[i]=inff,vis[i]=0,cnt[i]=0;
d[0]=0;
queue<int> q;
q.push(0);
while(!q.empty())
{
int u=q.front();q.pop();
vis[u]=0;
for(int i=head[u];~i;i=edge[i].p)
{
int v=edge[i].to;
if(d[v]>edge[i].val+d[u])
{
d[v]=edge[i].val+d[u];
if(!vis[v])
{
vis[v]=1;
if(++cnt[v]>n)
{
cout<<"CONFLICT"<<endl;
return ;
}
q.push(v);
}
}
}
}
if(fg) cout<<"UNCERTAIN"<<endl;
else cout<<"OK"<<endl;
}
int main()
{
while(cin>>n>>m)
{
int x,y;char ch;
init();
for(int i=1;i<=m;i++)
{
scanf("%d %c %d",&x,&ch,&y);
x++,y++;
meger(x,y);
if(ch=='=') add(y,x,0),add(x,y,0);
else if(ch=='>') add(x,y,-1);
else if(ch=='<') add(y,x,-1);
}
for(int i=1;i<=n;i++) add(0,i,0);
int sum=0;
fg=0;
for(int i=1;i<=n;i++)
if(i==find(i)) sum++;
if(sum>1) fg=1;
spfa();
}
}