POJ 3683 - Priest John's Busiest Day 构图2-sat..感觉求方案不太正确..要改进..
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题意:
在一个乡村有一个牧师John...有天很多人要举行婚礼..需要John来进行祷告..祷告必须要在婚礼开始时到开始后D时间段..或者到婚礼结束的最后D时间段..现在给出每个婚礼开始和结束的时间..以及每场婚礼需要祷告的时间长度...问所有婚礼是否能顺利进行..如果可以..输出任意一种牧师John祷告的时间方案...
题解:
对于一场婚礼.要么是前面的一段..要么是后面的一段..符合2-sat的基本模型..然后根据时间冲突的关系构造有向边...用tarjan判断是否有可行解..最后dfs染色找到答案..话说我越来越觉得我染色的那部分有bug...就算几个题都切过去了...但真的有很大问题!!!
Program:
#include<iostream>#include<stdio.h>#include<cmath>#include<queue>#include<stack>#include<string.h>#include<map>#include<set>#include<algorithm>#define oo 1000000007#define MAXN 1005<<1#define MAXM 1000005<<1#define ll long longusing namespace std; struct couple{ int t1,t2,d; }C[MAXN];struct node{ int x,y,next; }line[MAXM];set<int> T[MAXN];int Lnum,_next[MAXN],dfn[MAXN],low[MAXN],tp[MAXN],tpnum,DfsIndex,color[MAXN];bool instack[MAXN],d[MAXN];stack<int> mystack;void addline(int x,int y){ line[++Lnum].next=_next[x],_next[x]=Lnum; line[Lnum].x=x,line[Lnum].y=y;}bool legal(int s1,int t1,int s2,int t2){ if (s2<=s1 && s1<t2) return false; if (s2<t1 && t1<=t2) return false; if (s1<=s2 && s2<t1) return false; if (s1<t2 && t2<=t1) return false; return true;}void tarjan(int x){ int y,k; dfn[x]=low[x]=++DfsIndex; instack[x]=true,mystack.push(x); for (k=_next[x];k;k=line[k].next) { y=line[k].y; if (!dfn[y]) { tarjan(y); low[x]=min(low[x],low[y]); }else if (instack[y]) low[x]=min(low[x],dfn[y]); } if (low[x]==dfn[x]) { tpnum++; do { x=mystack.top(); mystack.pop(); instack[x]=false; tp[x]=tpnum; }while (low[x]!=dfn[x]); }}bool judge(int n){ for (int i=0;i<n;i++) if (tp[i<<1]==tp[i<<1|1]) return false; return true;}void dfs(int x){ if (color[x]) return; color[x]=1; for(set<int>::iterator i=T[x].begin();i!=T[x].end();i++) color[*i]=2; for (int k=_next[x];k;k=line[k].next) dfs(line[k].y); }int main(){ int N,i,j; scanf("%d",&N); for (i=0;i<N;i++) { int h,m; scanf("%d:%d",&h,&m); C[i].t1=h*60+m; scanf("%d:%d%d",&h,&m,&C[i].d); C[i].t2=h*60+m; } memset(_next,0,sizeof(_next)); Lnum=0; for (i=0;i<N;i++) for (j=i+1;j<N;j++) { if (!legal(C[i].t1,C[i].t1+C[i].d,C[j].t1,C[j].t1+C[j].d)) addline(i<<1,j<<1|1),addline(j<<1,i<<1|1); if (!legal(C[i].t1,C[i].t1+C[i].d,C[j].t2-C[j].d,C[j].t2)) addline(i<<1,j<<1),addline(j<<1|1,i<<1|1); if (!legal(C[i].t2-C[i].d,C[i].t2,C[j].t1,C[j].t1+C[j].d)) addline(i<<1|1,j<<1|1),addline(j<<1,i<<1); if (!legal(C[i].t2-C[i].d,C[i].t2,C[j].t2-C[j].d,C[j].t2)) addline(i<<1|1,j<<1),addline(j<<1|1,i<<1); } memset(dfn,0,sizeof(dfn)); memset(instack,false,sizeof(instack)); while (!mystack.empty()) mystack.pop(); DfsIndex=tpnum=0; for (i=0;i<(N<<1);i++) if (!dfn[i]) tarjan(i); if (!judge(N)) printf("NO\n"); else { printf("YES\n"); int temp=Lnum; Lnum=0; memset(_next,0,sizeof(_next)); memset(d,true,sizeof(d)); for (i=1;i<=tpnum;i++) T[i].clear(); for (i=0;i<(N<<1);i++) T[tp[i]].insert(tp[i^1]); for (i=1;i<=temp;i++) { int x=tp[line[i].x],y=tp[line[i].y]; if (x==y) continue; addline(x,y),d[y]=false; } do { memset(color,0,sizeof(color)); for (i=1;i<=tpnum;i++) if (!color[i]) dfs(i); for (i=0;i<N;i++) if (color[tp[i<<1]]==color[tp[i<<1|1]]) break; }while (i!=N); for (i=0;i<N;i++) { int h,m; if (color[tp[i<<1]]==1) { h=C[i].t1/60,m=C[i].t1%60; if (h<10) printf("0%d:",h); else printf("%d:",h); if (m<10) printf("0%d ",m); else printf("%d ",m); h=(C[i].t1+C[i].d)/60,m=(C[i].t1+C[i].d)%60; if (h<10) printf("0%d:",h); else printf("%d:",h); if (m<10) printf("0%d\n",m); else printf("%d\n",m); }else { h=(C[i].t2-C[i].d)/60,m=(C[i].t2-C[i].d)%60; if (h<10) printf("0%d:",h); else printf("%d:",h); if (m<10) printf("0%d ",m); else printf("%d ",m); h=C[i].t2/60,m=C[i].t2%60; if (h<10) printf("0%d:",h); else printf("%d:",h); if (m<10) printf("0%d\n",m); else printf("%d\n",m); } } } return 0;}
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