poj 3683 Priest John's Busiest Day 2-SAT
| Time Limit: 2000MS | Memory Limit: 65536K | |||
| Total Submissions: 7055 | Accepted: 2405 | 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
Source
祝福只能在婚礼的开头或结尾，枚举所有的婚礼对，判断开头结尾时间是否重叠，建图2-SAT即可。
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/** head-file **/
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <algorithm>
/** define-for **/
#define REP(i, n) for (int i=0;i<int(n);++i)
#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)
#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)
#define REP_1(i, n) for (int i=1;i<=int(n);++i)
#define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)
#define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)
#define REP_N(i, n) for (i=0;i<int(n);++i)
#define FOR_N(i, a, b) for (i=int(a);i<int(b);++i)
#define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i)
#define REP_1_N(i, n) for (i=1;i<=int(n);++i)
#define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i)
#define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i)
/** define-useful **/
#define clr(x,a) memset(x,a,sizeof(x))
#define sz(x) int(x.size())
#define see(x) cerr<<#x<<" "<<x<<endl
#define se(x) cerr<<" "<<x
#define pb push_back
#define mp make_pair
/** test **/
#define Display(A, n, m) { \
REP(i, n){ \
REP(j, m) cout << A[i][j] << " "; \
cout << endl; \
} \
}
#define Display_1(A, n, m) { \
REP_1(i, n){ \
REP_1(j, m) cout << A[i][j] << " "; \
cout << endl; \
} \
}
using namespace std;
/** typedef **/
typedef long long LL;
/** Add - On **/
const int direct4[4][2]={ {0,1},{1,0},{0,-1},{-1,0} };
const int direct8[8][2]={ {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int direct3[6][3]={ {1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1} };
const int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long INFF = 1LL << 60;
const double EPS = 1e-9;
const double OO = 1e15;
const double PI = acos(-1.0); //M_PI;
const int maxn=2111;
const int maxm=2111111;
int n,m;
struct EDGENODE{
int to;
int next;
};
struct TWO_SAT{
int head[maxn*2];
EDGENODE edges[maxm*2];
int edge;
int n;
bool mark[maxn*2];
int S[maxn*2],c;
void init(int n){
this->n=n;
clr(mark,0);
clr(head,-1);
edge=0;
}
void addedge(int u,int v){
edges[edge].to=v,edges[edge].next=head[u],head[u]=edge++;
}
// x = xval or y = yval
void add_clause(int x,int xval,int y,int yval){
x=x*2+xval;
y=y*2+yval;
addedge(x^1,y);
addedge(y^1,x);
}
void add_con(int x,int xval){
x=x*2+xval;
addedge(x^1,x);
}
bool dfs(int x){
if (mark[x^1]) return false;
if (mark[x]) return true;
mark[x]=true;
S[c++]=x;
for (int i=head[x];i!=-1;i=edges[i].next)
if (!dfs(edges[i].to)) return false;
return true;
}
bool solve(){
for (int i=0;i<n*2;i+=2)
if (!mark[i]&&!mark[i+1]){
c=0;
if (!dfs(i)){
while (c>0) mark[S[--c]]=false;
if (!dfs(i+1)) return false;
}
}
return true;
}
}TwoSAT;
struct TimeData{
int hh,mm;
TimeData(){}
TimeData(int h,int m){
hh=h,mm=m;
}
TimeData(char *s){
sscanf(s,"%d:%d",&hh,&mm);
}
void judge(){
while (mm<0){
hh--;
mm+=60;
}
if (mm>=60){
hh+=mm/60;
mm%=60;
}
}
friend TimeData operator+(const TimeData& a,const TimeData& b){
TimeData c;
c.hh=a.hh+b.hh;
c.mm=a.mm+b.mm;
c.judge();
return c;
}
friend TimeData operator+(const TimeData& a,const int& b){
TimeData c;
c=a;
c.mm+=b;
c.judge();
return c;
}
friend TimeData operator-(const TimeData& a,const int& b){
TimeData c;
c=a;
c.mm-=b;
c.judge();
return c;
}
friend bool operator<(const TimeData& a,const TimeData& b){
if (a.hh==b.hh){
return a.mm<b.mm;
}
return a.hh<b.hh;
}
friend bool operator<=(const TimeData& a,const TimeData& b){
if (a.hh==b.hh){
return a.mm<=b.mm;
}
return a.hh<b.hh;
}
friend bool operator>(const TimeData& a,const TimeData& b){
if (a.hh==b.hh){
return a.mm>b.mm;
}
return a.hh>b.hh;
}
friend bool operator>=(const TimeData& a,const TimeData& b){
if (a.hh==b.hh){
return a.mm>=b.mm;
}
return a.hh>b.hh;
}
friend bool operator==(const TimeData& a,const TimeData& b){
return (a.hh==b.hh&&a.mm==b.mm);
}
void output(){
printf("%2.2d:%2.2d",hh,mm);
}
};
struct Wedding{
TimeData s,t;
int d;
TimeData sleft(){
return s;
}
TimeData sright(){
return s+d;
}
TimeData tleft(){
return t-d;
}
TimeData tright(){
return t;
}
void output(){
s.output();
printf(" ");
t.output();
printf(" %d\n",d);
}
}a[maxn];
int main()
{
char ss[11];
while (~scanf("%d",&n))
{
TwoSAT.init(n);
REP(i,n){
scanf("%s",ss);
a[i].s=TimeData(ss);
scanf("%s",ss);
a[i].t=TimeData(ss);
scanf("%d",&a[i].d);
}
//REP(i,n) a[i].output();
REP(i,n){
FOR(j,i+1,n){
if (!(a[i].sright()<=a[j].sleft()||a[i].sleft()>=a[j].sright())) TwoSAT.add_clause(i,1,j,1);
if (!(a[i].sright()<=a[j].tleft()||a[i].sleft()>=a[j].tright())) TwoSAT.add_clause(i,1,j,0);
if (!(a[i].tright()<=a[j].sleft()||a[i].tleft()>=a[j].sright())) TwoSAT.add_clause(i,0,j,1);
if (!(a[i].tright()<=a[j].tleft()||a[i].tleft()>=a[j].tright())) TwoSAT.add_clause(i,0,j,0);
}
}
if (TwoSAT.solve()){
puts("YES");
for (int i=0;i<n;i++)
{
if (TwoSAT.mark[i*2])
{
a[i].sleft().output();
printf(" ");
a[i].sright().output();
printf("\n");
}
else
{
a[i].tleft().output();
printf(" ");
a[i].tright().output();
printf("\n");
}
}
}else{
puts("NO");
}
}
return 0;
}
