|poj 3683|2-SAT|Priest John's Busiest Day

/*
poj 3683
2-SAT
教训：
1、区间重合的判定
2、想要输出数最少两位可以用"%.2d"输出
例如数是8，但想输出08，就可以用，如果是14，则还是输出14
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define ms(i,j) memset(i, j, sizeof i);

const int MAXN = 1000 + 5;

struct inv
{
int l, r, time;
}in[MAXN];

struct TwoSAT
{
vector<int> G[MAXN*2];
int S[MAXN*2];
bool mark[MAXN*2];
int n, c;
void init(int n)
{
this->n = n;
for (int i=0;i<=n*2;i++)
{
G[i].clear();
mark[i] = false;
}
}
void ins(int u, int v)
{
G[u].push_back(v);
}
int dfs(int x)
{
if (mark[x^1]) return false;
if (mark[x]) return true;
mark[x] = true;
S[++c] = x;
for (int i=0;i<G[x].size();i++)
{
if (!dfs(G[x][i])) return false;
}
return true;
}
int solve()
{
for (int i=0;i<2*n;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;
}
}ts;

int n;
int fight(int x1, int y1, int x2, int y2)
{
if (x2>=x1&&x2<y1) return true;
if (y2<=y1&&x1<y2) return true;
return false;
}
void init()
{
for (int i=0;i<n;i++)
{
int h1, h2, m1, m2;
scanf("%d:%d %d:%d %d", &h1, &m1, &h2, &m2, &in[i].time);
in[i].l = h1*60+m1;
in[i].r = h2*60+m2;
}
}
void solve()//2*x is left, 2*x+1 is right(~)
{
ts.init(n);
for (int i=0;i<n;i++)
for (int j=0;j<n;j++)
if (i!=j)
{
int &x1 = in[i].l, &y1 = in[i].r, &x2 = in[j].l, &y2 = in[j].r, &t1 = in[i].time, &t2 = in[j].time;
if (fight(x1, x1+t1, x2, x2+t2))//l l
{
ts.ins(2*i, 2*j+1);
ts.ins(2*j, 2*i+1);
}
if (fight(x1, x1+t1, y2-t2, y2))//l r
{
ts.ins(2*i, 2*j);
ts.ins(2*j+1, 2*i+1);
}
if (fight(y1-t1, y1, y2-t2, y2))//r r
{
ts.ins(2*i+1, 2*j);
ts.ins(2*j+1, 2*i);
}
if (fight(y1-t1, y1, x2, x2+t2))//r l
{
ts.ins(2*j, 2*i);
ts.ins(2*i+1, 2*j+1);
}
}
if (ts.solve())
{
printf("YES\n");
for (int i=0;i<n;i++)
{
if (ts.mark[i*2])//l
{
int tot = in[i].l + in[i].time;
int h1 = in[i].l / 60, m1 = in[i].l % 60;
int h2 = tot / 60, m2 = tot % 60;
printf("%.2d:%.2d %.2d:%.2d\n", h1, m1, h2, m2);
}
if (ts.mark[i*2+1])//r
{
int tot = in[i].r - in[i].time;
int h1 = in[i].r / 60, m1 = in[i].r % 60;
int h2 = tot / 60, m2 = tot % 60;
printf("%.2d:%.2d %.2d:%.2d\n", h2, m2, h1, m1);
}
}
} else printf("NO\n");
}
int main()
{
while (scanf("%d", &n)==1)
{
init();
solve();
}
return 0;
}