2-sat 输出任意一组可行解&拓扑排序+缩点 poj3683
| Time Limit: 2000MS | Memory Limit: 65536K | |||
| Total Submissions: 8170 | Accepted: 2784 | Special Judge | ||
Description
John is the only priest in his town. September 1st is the John's busiest day in a year because there is an old legend in the town that the couple who get married on that day will be forever blessed by the God of Love. This year N couples plan to get married on the blessed day. The i-th couple plan to hold their wedding from time Si to time Ti. According to the traditions in the town, there must be a special ceremony on which the couple stand before the priest and accept blessings. The i-th couple need Di minutes to finish this ceremony. Moreover, this ceremony must be either at the beginning or the ending of the wedding (i.e. it must be either from Si to Si + Di, or from Ti - Di to Ti). Could you tell John how to arrange his schedule so that he can present at every special ceremonies of the weddings.
Note that John can not be present at two weddings simultaneously.
Input
The first line contains a integer N ( 1 ≤ N ≤ 1000).
The next N lines contain the Si, Ti and Di. Si and Ti are in the format of hh:mm.
Output
The first line of output contains "YES" or "NO" indicating whether John can be present at every special ceremony. If it is "YES", output another N lines describing the staring time and finishing time of all the ceremonies.
Sample Input
2 08:00 09:00 30 08:15 09:00 20
Sample Output
YES 08:00 08:30 08:40 09:00
方法一：tarjan缩点+拓扑排序
#include"stdio.h"
#include"string.h"
#include"stdlib.h"
#include"queue"
#include"algorithm"
#include"string.h"
#include"string"
#include"vector"
#include"stack"
#include"map"
#define inf 0x3f3f3f3f
#define M 2009
using namespace std;
struct node
{
int u,v,next;
}edge[M*M*5];
stack<int>q;
struct EDGE
{
int v;
EDGE(int vv)
{
v=vv;
}
};
vector<EDGE>Edge[M];
int t,head[M],low[M],dfn[M],belong[M],num,index,use[M],n,degree[M],Color[M],fp[M];
void init()
{
t=0;
memset(head,-1,sizeof(head));
}
void add(int u,int v)
{
edge[t].u=u;
edge[t].v=v;
edge[t].next=head[u];
head[u]=t++;
}
void tarjan(int u)
{
low[u]=dfn[u]=++index;
q.push(u);
use[u]=1;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(!dfn[v])
{
tarjan(v);
low[u]=min(low[u],low[v]);
}
else if(use[v])
low[u]=min(low[u],dfn[v]);
}
if(low[u]==dfn[u])
{
num++;
int vv;
do
{
vv=q.top();
q.pop();
use[vv]=0;
belong[vv]=num;
}while(vv!=u);
}
}
int psq(int n)
{
int i;
num=index=0;
memset(use,0,sizeof(use));
memset(dfn,0,sizeof(dfn));
for(i=1;i<=2*n;i++)
if(!dfn[i])
tarjan(i);
for(i=1;i<=n;i++)
if(belong[i]==belong[i+n])
return 0;
return 1;
}
struct Time
{
int l,r;
}time[M];
int ok(Time a,Time b)
{
if(a.r<=b.l||b.r<=a.l)
return 0;
return 1;
}
int op(int u)
{
if(u<=n)
return n+u;
else
return u-n;
}
int main()
{
int i,j;
while(scanf("%d",&n)!=-1)
{
for(i=1;i<=n;i++)
{
int h1,h2,m1,m2,d;
scanf("%d:%d %d:%d %d",&h1,&m1,&h2,&m2,&d);
time[i].l=h1*60+m1;
time[i].r=h1*60+m1+d;
time[i+n].l=h2*60+m2-d;
time[i+n].r=h2*60+m2;
}
init();
for(i=1;i<=n;i++)
{
for(j=i+1;j<=n;j++)
{
if(ok(time[i],time[j]))
{
add(i,j+n);
add(j,i+n);
}
if(ok(time[i],time[j+n]))
{
add(i,j);
add(j+n,i+n);
}
if(ok(time[i+n],time[j]))
{
add(i+n,j+n);
add(j,i);
}
if(ok(time[i+n],time[j+n]))
{
add(i+n,j);
add(j+n,i);
}
}
}
if(psq(n))//拓扑排序输出可行解
{
printf("YES\n");
memset(degree,0,sizeof(degree));//缩点的入度
memset(Color,0,sizeof(Color));//染色
for(i=0;i<t;i++)
{
int u=edge[i].u;
int v=edge[i].v;
fp[belong[u]]=belong[op(u)];//记录当前点所在的联通块与对应点所在联通块的相互影射
fp[belong[op(u)]]=belong[u];
if(belong[u]!=belong[v])
{
Edge[belong[v]].push_back(EDGE(belong[u]));//缩点建图，若不在同一个连通块则建立反边
degree[belong[u]]++;
}
}
queue<int>q;
for(i=1;i<=num;i++)
if(degree[i]==0)q.push(i);//入度为0的点入队
while(!q.empty())
{
int u=q.front();
q.pop();
if(Color[u]==0)//对未着色的点着色同时把对立点所在连通块着为相反的颜色
{
Color[u]=1;
Color[fp[u]]=-1;
}
for(i=0;i<(int)Edge[u].size();i++)
{
int v=Edge[u][i].v;
if(--degree[v]==0)
q.push(v);
}
}
for(i=1;i<=n;i++)
{
if(Color[belong[i]]==1)//连通块标记为1的是可行解
printf("%02d:%02d %02d:%02d\n",time[i].l/60,time[i].l%60,time[i].r/60,time[i].r%60);
else
printf("%02d:%02d %02d:%02d\n",time[i+n].l/60,time[i+n].l%60,time[i+n].r/60,time[i+n].r%60);
}
for(i=1;i<=num;i++)
Edge[i].clear();
}
else
printf("NO\n");
}
}
方法二：dfs+枚举（字典序最小）#include"stdio.h" #include"string.h" #include"stdlib.h" #include"queue" #include"algorithm" #include"string.h" #include"string" #include"vector" #include"stack" #include"map" #define inf 0x3f3f3f3f #define M 2009 using namespace std; struct node { int u,v,next; }edge[M*M*5]; stack<int>q; struct EDGE { int v; EDGE(int vv) { v=vv; } }; vector<EDGE>Edge[M]; int t,head[M],low[M],dfn[M],belong[M],num,index,use[M],color[M],s[M],cnt,n,degree[M],Color[M],fp[M]; void init() { t=0; memset(head,-1,sizeof(head)); } void add(int u,int v) { edge[t].u=u; edge[t].v=v; edge[t].next=head[u]; head[u]=t++; } struct Time { int l,r; }time[M]; int ok(Time a,Time b) { if(a.r<=b.l||b.r<=a.l) return 0; return 1; } int op(int u) { if(u<=n) return n+u; else return u-n; } int dfs(int u) { if(color[u]==1) return 1; if(color[u]==-1) return 0; s[cnt++]=u; color[u]=1; color[op(u)]=-1; for(int i=head[u];i!=-1;i=edge[i].next) { int v=edge[i].v; if(!dfs(v)) return 0; } return 1; } int slove(int n) { int i,j; memset(color,0,sizeof(color)); for(i=1;i<=2*n;i++) { if(color[i])continue; cnt=0; if(!dfs(i)) { for(j=0;j<cnt;j++) color[s[j]]=color[op(s[j])]=0; if(!dfs(op(i))) return 0; } } return 1; } int main() { int i,j; while(scanf("%d",&n)!=-1) { for(i=1;i<=n;i++) { int h1,h2,m1,m2,d; scanf("%d:%d %d:%d %d",&h1,&m1,&h2,&m2,&d); time[i].l=h1*60+m1; time[i].r=h1*60+m1+d; time[i+n].l=h2*60+m2-d; time[i+n].r=h2*60+m2; } init(); for(i=1;i<=n;i++) { for(j=i+1;j<=n;j++) { if(ok(time[i],time[j])) { add(i,j+n); add(j,i+n); } if(ok(time[i],time[j+n])) { add(i,j); add(j+n,i+n); } if(ok(time[i+n],time[j])) { add(i+n,j+n); add(j,i); } if(ok(time[i+n],time[j+n])) { add(i+n,j); add(j+n,i); } } } if(slove(n)) { printf("YES\n"); for(i=1;i<=n;i++) { if(color[i]==1) printf("%02d:%02d %02d:%02d\n",time[i].l/60,time[i].l%60,time[i].r/60,time[i].r%60); if(color[i+n]==1) printf("%02d:%02d %02d:%02d\n",time[i+n].l/60,time[i+n].l%60,time[i+n].r/60,time[i+n].r%60); } for(i=1;i<=num;i++) Edge[i].clear(); } else printf("NO\n"); } }
