pku 3683 Priest John's Busiest Day 2-sat判断有误解+输出可行解
http://poj.org/problem?id=3683
题意：
一个教父，在一天中要给n对新婚夫妇举行婚礼。已知每对夫妇举行婚礼的起始时间Si和终止时间Ti ,教父送祝福的时间要么在Si->Si +Di 要么在Ti - Di->Ti。问在这一天中，教父如何安排才能对着n对新人都送去祝福？
思路：
把每对新人的婚礼的起始时间和结束时间加减Di后分成两个对立的点，然后检查每个点（这里每个点表示一个时间段，教父在这一事件段里送上祝福）是否存在区间相交，如果i与j存在相交则
i->j^1建边。然后将缩点后的图重新反向建图，利用拓扑排序，输出解。
//#pragma comment(linker,"/STACK:327680000,327680000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>
#define CL(arr, val) memset(arr, val, sizeof(arr))
#define ll long long
#define inf 0x7f7f7f7f
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define ll long long
#define L(x) (x) << 1
#define R(x) (x) << 1 | 1
#define MID(l, r) (l + r) >> 1
#define Min(x, y) (x) < (y) ? (x) : (y)
#define Max(x, y) (x) < (y) ? (y) : (x)
#define E(x) (1 << (x))
#define iabs(x) (x) < 0 ? -(x) : (x)
#define OUT(x) printf("%I64d\n", x)
#define lowbit(x) (x)&(-x)
#define Read() freopen("din.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);
#define N 2007
#define M 1000007
using namespace std;
struct node
{
int v,u;
int next;
}g[N*N],rg[N*N];
int head[N],ct;
int H[N],ht;
int dfn[N],low[N];
int belong[N],stk[N];
bool isn[N];
int col[N],ind[N],opp[N];
int idx,cnt,top;
struct time
{
int s,e;
}tm[N];
int n;
void add(int u,int v)
{
g[ct].u = u;
g[ct].v = v;
g[ct].next = head[u];
head[u] = ct++;
}
void build()
{
int i,j;
CL(head,-1); ct = 0;
for (i = 0; i < 2*n; ++i)
{
low[i] = dfn[i] = -1;
belong[i] = 0;
col[i] = 0; ind[i] = 0;
isn[i] = false;
opp[i] = 0;
for (j = 0; j < 2*n; ++j)
{
if (i == j || (i^1) == j) continue;
if (tm[i].s < tm[j].e && tm[i].e > tm[j].s)
{
add(i,j^1);
}
}
}
idx = top = cnt = 0;
}
void radd(int u,int v)
{
rg[ht].v = v;
rg[ht].next = H[u];
H[u] = ht++;
}
void rbuild()
{
CL(H,-1); ht = 0;
for (int i = 0; i < ct; ++i)
{
if (belong[g[i].u] != belong[g[i].v])
{
radd(belong[g[i].v],belong[g[i].u]);
ind[belong[g[i].u]]++;
}
}
}
void topsort()
{
int i;
queue<int>q;
for (i = 1; i <= cnt; ++i)
{
if (ind[i] == 0) q.push(i);
}
while (!q.empty())
{
int u = q.front(); q.pop();
if (col[u] == 0)
{
col[u] = 1;
col[opp[u]] = -1;
}
for (i = H[u]; i != -1; i = rg[i].next)
{
int v = rg[i].v;
ind[v]--;
if (ind[v] == 0) q.push(v);
}
}
}
void tarjan(int u)
{
int i,j;
dfn[u] = low[u] = ++idx;
isn[u] = true;
stk[++top] = u;
for (i = head[u]; i != -1; i = g[i].next)
{
int v = g[i].v;
if (dfn[v] == -1)
{
tarjan(v);
low[u] = min(low[u],low[v]);
}
else if (isn[v])
{
low[u] = min(low[u],dfn[v]);
}
}
if (dfn[u] == low[u])
{
cnt++;
do
{
j = stk[top--];
isn[j] = false;
belong[j] = cnt;
} while (j != u);
}
}
void solve()
{
int i;
for (i = 0; i < 2*n; ++i)
{
if (dfn[i] == -1) tarjan(i);
}
bool flag = false;
for (i = 0; i < n; ++i)
{
if (belong[2*i] == belong[2*i + 1])
{
flag = true;
break;
}
opp[belong[2*i]] = belong[2*i + 1];//记录每个点的冲突点
opp[belong[2*i + 1]] = belong[2*i];
}
if (flag) printf("NO\n");
else
{
printf("YES\n");
rbuild();
topsort();
for (i = 0; i < 2*n; i += 2)
{
if (col[belong[i]] == 1)
{
printf("%02d:%02d %02d:%02d\n",tm[i].s/60,tm[i].s%60,tm[i].e/60,tm[i].e%60);
}
else
{
printf("%02d:%02d %02d:%02d\n",tm[i + 1].s/60,tm[i + 1].s%60,tm[i + 1].e/60,tm[i + 1].e%60);
}
}
}
}
int main()
{
// Read();
int i;
int h1,m1,h2,m2,len;
while (~scanf("%d",&n))
{
for (i = 0; i < n; ++i)
{
scanf("%d:%d %d:%d %d",&h1,&m1,&h2,&m2,&len);
//printf("%d %d %d %d %d",h1,m1,h2,m2,len);
tm[2*i].s = h1*60 + m1;
tm[2*i].e = tm[2*i].s + len;
tm[2*i + 1].e = h2*60 + m2;
tm[2*i + 1].s = tm[2*i + 1].e - len;
}
build();
solve();
}
return 0;
}