Description
Conflicts are everywhere in the world, from the young
to the elderly, from families to countries. Conflicts cause quarrels, fights or
even wars. How wonderful the world will be if all conflicts can be
eliminated.
Edward contributes his lifetime to invent a ‘Conflict Resolution
Terminal‘ and he has finally succeeded. This magic item has the ability to
eliminate all the conflicts. It works like this:
If any two people have
conflict, they should simply put their hands into the ‘Conflict Resolution
Terminal‘ (which is simply a plastic tube). Then they play ‘Rock, Paper and
Scissors‘ in it. After they have decided what they will play, the tube should be
opened and no one will have the chance to change. Finally, the winner have the
right to rule and the loser should obey it. Conflict Eliminated!
But the game
is not that fair, because people may be following some patterns when they play,
and if the pattern is founded by others, the others will win
definitely.
Alice and Bob always have conflicts with each other so they use
the ‘Conflict Resolution Terminal‘ a lot. Sadly for Bob, Alice found his pattern
and can predict how Bob plays precisely. She is very kind that doesn‘t want to
take advantage of that. So she tells Bob about it and they come up with a new
way of eliminate the conflict:
They will play the ‘Rock, Paper and Scissors‘
for N round. Bob will set up some restricts on Alice.
But the restrict can
only be in the form of "you must play the same (or different) on the ith and jth
rounds". If Alice loses in any round or break any of the rules she loses,
otherwise she wins.
Will Alice have a chance to win?
Input
The first line contains an integer T(1 <= T <=
50), indicating the number of test cases.
Each test case contains several
lines.
The first line contains two integers N,M(1 <= N <= 10000, 1
<= M <= 10000), representing how many round they will play and how many
restricts are there for Alice.
The next line contains N integers
B1,B2, ...,BN, where
Bi represents what item Bob will play in the
ith round. 1 represents Rock, 2 represents Paper, 3 represents
Scissors.
The following M lines each contains three integers A,B,K(1 <=
A,B <= N,K = 0 or 1) represent a restrict for Alice. If K equals 0, Alice
must play the same on Ath and Bth round. If K
equals 1, she must play different items on Ath and Bthround.
Output
For each test case in the input, print one line:
"Case #X: Y", where X is the test case number (starting with 1) and Y is "yes"
or "no" represents whether Alice has a chance to win.
Sample Input
2 3 3 1 1 1 1 2 1 1 3 1 2 3 1 5 5 1 2 3 2 1 1 2 1 1 3 1 1 4 1 1 5 1 2 3
0
Sample Output
Case #1: no Case #2: yes
Hint
‘Rock, Paper and Scissors‘ is a game which played by two person. They should play Rock, Paper or Scissors by their hands at the same time.
Rock defeats scissors, scissors defeats paper and paper defeats rock. If two people play the same item, the game is tied..
这个题有意思了啊，看着感觉是3-SAT，可是分析一下是2-SAT，简单，所谓的2-SAT不就是建立一个有向图，再求SCC，以就是强连通分量，再看看同一个点的两种状态在不在同一个SCC上，简单题，
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define inf 0x0f0f0f0f
using namespace std;
int finger[10000+10][2];
struct SCC
{
static const int maxn=20000+10;
vector<int>group[maxn],scc[maxn];
int pre[maxn],lowlink[maxn],sccno[maxn],dfs_clock,scc_cnt,n,m;
stack<int>S;void init()
{
for (int i=0;i<=n;i++) group[i].clear();
}void addedge(int from,int to)
{
group[from].push_back(to);
}void dfs(int u)
{
pre[u]=lowlink[u]=++dfs_clock;
S.push(u);
for (int i=0;i<group[u].size();i++)
{
int v=group[u][i];
if (!pre[v])
{
dfs(v);
lowlink[u]=min(lowlink[u],lowlink[v]);
}
else if (!sccno[v])
{
lowlink[u]=min(lowlink[u],pre[v]);
}
}
if (lowlink[u]==pre[u])
{
scc_cnt++;
scc[scc_cnt].clear();
while (1)
{
int x=S.top();
S.pop();
scc[scc_cnt].push_back(x);
sccno[x]=scc_cnt;
if (x==u) break;
}
}
}void find_scc()
{
dfs_clock=scc_cnt=0;
memset(pre,0,sizeof(pre));
memset(sccno,0,sizeof(sccno));
for (int i=1;i<=n;i++)
if (!pre[i]) dfs(i);
}
};SCC Conflict;
int main()
{
int T,N,M,x,y,c,cas=0;
scanf("%d",&T);
while (T--)
{
cas++;
scanf("%d%d",&N,&M);
Conflict.n=N*2;
Conflict.init();
for (int i=0;i<N;i++)
{
scanf("%d",&x);
if (x==1)
{
finger[i][0]=1;
finger[i][1]=2;
}
if (x==2)
{
finger[i][0]=2;
finger[i][1]=3;
}
if (x==3)
{
finger[i][0]=3;
finger[i][1]=1;
}
}
for (int i=1;i<=M;i++)
{
scanf("%d%d%d",&x,&y,&c);
x--;y--;
if (c==0)
{
for (int j=0;j<2;j++)
{
for (int k=0;k<2;k++)
if (finger[x][j]!=finger[y][k])
{
Conflict.addedge(x*2+j,(y*2+k)^1);
Conflict.addedge(y*2+k,(x*2+j)^1);
}
}
}
if (c==1)
{
for (int j=0;j<2;j++)
{
for (int k=0;k<2;k++)
if (finger[x][j]==finger[y][k])
{
Conflict.addedge(x*2+j,(y*2+k)^1);
Conflict.addedge(y*2+k,(x*2+j)^1);
}
}
}
}
Conflict.find_scc();
bool ans=true;
for (int i=0;i<N;i+=2)
{
if (Conflict.sccno[i]==Conflict.sccno[i+1])
{
ans=false;
break;
}
}
if (ans) printf("Case #%d: yes\n",cas);
else printf("Case #%d: no\n",cas);
}
return 0;
}
作者 chensunrise
hdu 4115 Eliminate the Conflict,布布扣,bubuko.com