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Priest John's Busiest Day##
Time Limit: 2 Sec
Memory Limit: 64 MBDescription###
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
题目地址： poj 3683
题目大意：
题目已经很简洁了>_<
题解：
先判是否有解
然后对于 A 和 'A
它们所在的联通块是不能同时选的
重新建图
拓扑一下搞出顺序
就完了
AC代码
//poj3683
#include <cstdio>
#include <algorithm>
using namespace std;
const int N=1e3+5;
int n,cnt,_cnt;
int S1[N],T1[N],S2[N],T2[N],last[N<<1],_last[N<<1];
int op[N<<1];
inline int read(){
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
bool judge(int S1,int T1,int S2,int T2){
if(T1<=S2 || S1>=T2)return 0;
return 1;
}
struct edge{
int to,next;
}e[(N*N)<<2],_e[(N*N)<<2];
void add_edge(int u,int v){
e[++cnt]=(edge){v,last[u]};last[u]=cnt;
}
int in[N<<1];
void _add_edge(int u,int v){
in[v]++;
_e[++_cnt]=(edge){v,_last[u]};_last[u]=_cnt;
}
int ind,top,scc;
int q[N<<1],dfn[N<<1],low[N<<1],col[N<<1];
bool inq[N<<1];
void Tarjan(int u){
dfn[u]=low[u]=++ind;
q[++top]=u;inq[u]=1;
for(int i=last[u];i;i=e[i].next){
int v=e[i].to;
if(!dfn[v]){
Tarjan(v);
low[u]=min(low[u],low[v]);
}
else if(inq[v])
low[u]=min(low[u],dfn[v]);
}
if(low[u]==dfn[u]){
int now=0;scc++;
while(now!=u){
now=q[top--];
col[now]=scc;
inq[now]=0;
}
}
}
int mark[N<<1];
void dfs(int u){
if(mark[u])return;
mark[u]=-1;
for(int i=_last[u];i;i=_e[i].next)
dfs(_e[i].to);
}
void topsort(){
top=0;
for(int i=1;i<=scc;i++)
if(!in[i])q[++top]=i;
while(top){
int u=q[top--];
if(mark[u])continue;
mark[u]=1;dfs(op[u]);
for(int i=_last[u];i;i=_e[i].next){
int v=_e[i].to;
in[v]--;
if(!in[v])q[++top]=v;
}
}
}
void print(int x){
printf("%.2d:",x/60);
printf("%.2d ",x%60);
}
int main(){
n=read();
for(int i=1;i<=n;i++){
S1[i]=read()*60+read();
T2[i]=read()*60+read();
int x=read();
T1[i]=S1[i]+x;
S2[i]=T2[i]-x;
}
for(int i=1;i<=n;i++)
for(int j=i+1;j<=n;j++){
if(judge(S1[i],T1[i],S1[j],T1[j])){
add_edge(i*2,j*2-1);
add_edge(j*2,i*2-1);
}
if(judge(S1[i],T1[i],S2[j],T2[j])){
add_edge(i*2,j*2);
add_edge(j*2-1,i*2-1);
}
if(judge(S2[i],T2[i],S1[j],T1[j])){
add_edge(i*2-1,j*2-1);
add_edge(j*2,i*2);
}
if(judge(S2[i],T2[i],S2[j],T2[j])){
add_edge(i*2-1,j*2);
add_edge(j*2-1,i*2);
}
}
for(int i=1;i<=n*2;i++)
if(!dfn[i])Tarjan(i);
for(int i=1;i<=n;i++)
if(col[i*2-1]==col[i*2]){
puts("NO");
return 0;
}
puts("YES");
for(int u=1;u<=2*n;u++)
for(int i=last[u];i;i=e[i].next)
if(col[u]!=col[e[i].to])
_add_edge(col[e[i].to],col[u]);
for(int i=1;i<=n;i++){
op[col[i*2]]=col[i*2-1];
op[col[i*2-1]]=col[i*2];
}
topsort();
for(int i=1;i<=n;i++)
if(mark[col[i*2]]==1)
print(S1[i]),print(T1[i]),puts("");
else print(S2[i]),print(T2[i]),puts("");
return 0;
}