POJ3683 Priest John's Busiest Day

原题链接

\(2-SAT\)模板题。
对于有时间重叠的婚礼转换成\(2-SAT\)的命题形式连边，用\(tarjan\)找强连通分量并判断，确定方案即可。
然而一道模板题，我因为数组开小了调了一晚上。。。

#include<cstdio>
using namespace std;
const int N = 2010;
const int M = 1e6 + 10;
struct dd {
    int x[2], y[2], z, S[2];
};
dd a[N >> 1];
int fi[N], di[M << 2], ne[M << 2], dfn[N], low[N], st[N], bl[N], l, ti, tp, SCC;
bool v[N];
inline int re()
{
    int x = 0;
    char c = getchar();
    bool p = 0;
    for (; c < '0' || c > '9'; c = getchar())
        p |= c == '-';
    for (; c >= '0' && c <= '9'; c = getchar())
        x = x * 10 + c - '0';
    return p ? -x : x;
}
inline void add(int x, int y)
{
    di[++l] = y;
    ne[l] = fi[x];
    fi[x] = l;
}
inline int minn(int x, int y)
{
    return x < y ? x : y;
}
void tarjan(int x)
{
    int i, y;
    dfn[x] = low[x] = ++ti;
    st[++tp] = x;
    v[x] = 1;
    for (i = fi[x]; i; i = ne[i])
    {
        y = di[i];
        if (!dfn[y])
        {
            tarjan(y);
            low[x] = minn(low[x], low[y]);
        }
        else
            if (v[y])
                low[x] = minn(low[x], dfn[y]);
    }
    if (!(dfn[x] ^ low[x]))
    {
        ++SCC;
        do
        {
            y = st[tp--];
            bl[y] = SCC;
            v[y] = 0;
        } while (x ^ y);
    }
}
void pr(int p, int i)
{
    printf("%02d:%02d ", a[i].x[p], a[i].y[p]);
    a[i].y[p] += a[i].z;
    a[i].x[p] += a[i].y[p] / 60;
    a[i].y[p] %= 60;
    printf("%02d:%02d\n", a[i].x[p], a[i].y[p]);
}
int main()
{
    int i, n, j, x, y, z;
    n = re();
    for (i = 1; i <= n; i++)
    {
        a[i].x[0] = re();
        a[i].y[0] = re();
        a[i].x[1] = re();
        a[i].y[1] = re();
        a[i].z = re();
        a[i].S[0] = 60 * a[i].x[0] + a[i].y[0];
        a[i].S[1] = 60 * a[i].x[1] + a[i].y[1] - a[i].z;
        while (a[i].y[1] < a[i].z)
        {
            a[i].x[1]--;
            a[i].y[1] += 60;
        }
        a[i].y[1] -= a[i].z;
    }
    for (i = 1; i < n; i++)
        for (j = i + 1; j <= n; j++)
            for (x = 0; x < 2; x++)
                for (y = 0; y < 2; y++)
                    if (!(a[i].S[x] + a[i].z <= a[j].S[y] || a[i].S[x] >= a[j].S[y] + a[j].z))
                    {
                        add(i + x * n, j + (y ^ 1) * n);
                        add(j + y * n, i + (x ^ 1) * n);
                    }
    for (i = 1; i <= (n << 1); i++)
        if (!dfn[i])
            tarjan(i);
    for (i = 1; i <= n; i++)
        if (!(bl[i] ^ bl[i + n]))
        {
            printf("NO");
            return 0;
        }
    printf("YES\n");
    for (i = 1; i <= n; i++)
    {
        if (bl[i] > bl[i + n])
            pr(1, i);
        else
            pr(0, i);
    }
    return 0;
}