题目大意:有n个问题,m个人来投票,没人最多投4票,问该怎样决定才能使每个人都有超过一半的票数被认可?
题目分析:2-SAT问题。如果某个人投的票数少于2,则这两票军被采纳,如果票数至少三票,则最多有一票可以不被采纳,这意味着这个人的投的任意两票之间有矛盾,是“二者取一”的关系。
代码如下:
# include<iostream> # include<cstdio> # include<vector> # include<queue> # include<cstring> # include<algorithm> using namespace std; # define LL long long # define REP(i,s,n) for(int i=s;i<n;++i) # define CLS(a,b) memset(a,b,sizeof(a)) # define CLL(a,b,n) fill(a,a+n,b) const int N=105; vector<int>G[N<<1]; int n,m,a[5],b[5],must[N+N],mark[N+N],flag[N],s[N+N],cnt; void add(int x,int u,int y,int v) { x=x*2+u; y=y*2+v; G[x^1].push_back(y); G[y^1].push_back(x); } void dfs1(int u) { must[u]=1; REP(i,0,G[u].size()) if(!must[G[u][i]]) dfs1(G[u][i]); } void clear() { CLS(mark,0); REP(i,0,n+n) mark[i]=must[i]; } bool dfs(int x) { if(mark[x^1]) return false; if(mark[x]) return true; mark[x]=1; s[cnt++]=x; REP(i,0,G[x].size()) if(!dfs(G[x][i])) return false; return true; } bool solve() { for(int i=0;i<n+n;i+=2){ if(mark[i]&&mark[i+1]) return false; if(!mark[i]&&!mark[i+1]){ cnt=0; if(!dfs(i)){ while(cnt>0) mark[s[--cnt]]=0; if(!dfs(i+1)) return false; } } } return true; } void init() { REP(i,0,n*2) G[i].clear(); CLS(must,0); CLS(mark,0); } int main() { int k,cas=0; char c; while(scanf("%d%d",&n,&m)&&(n+m)) { init(); CLS(flag,0); while(m--) { scanf("%d",&k); REP(i,0,k){ scanf("%d %c",&a[i],&c); --a[i]; if(c==‘y‘) b[i]=1; else b[i]=0; } if(k<=2){ REP(i,0,k) must[a[i]*2+b[i]]=1; }else REP(i,0,k) REP(j,i+1,k) add(a[i],b[i],a[j],b[j]); } REP(i,0,n+n) if(must[i]) dfs1(i); printf("Case %d: ",++cas); REP(i,0,n){ if(must[i*2]||must[i*2+1]) continue; clear(); mark[i*2]=1; bool yy1=solve(); clear(); mark[i*2+1]=1; bool yy2=solve(); if(yy1&&yy2) flag[i]=1; } clear(); if(!solve()){ printf("impossible\n"); continue; } REP(i,0,n){ if(flag[i]) printf("?"); else if(mark[i*2]) printf("n"); else printf("y"); } printf("\n"); } return 0; }
UVALive-4452 The Ministers' Major Mess (2-SAT)
时间: 2024-10-10 09:38:55