「JSOI2014」学生选课

「JSOI2014」学生选课

传送门
看到这题首先可以二分。
考虑对于当前的 \(mid\) 如何 \(\text{check}\)
我们用 \(f_{i,j}\) 来表示 \(i\) 对 \(j\) 的好感度排名，那么对于两个人 \(i\)，\(j\) 如果有 \(\max\{f_{i, j}, f_{j, i}\} > mid\) 那么显然这两个人是不能上同一个老师的课的。
而且每个人可以上的课只有两种，我们记为 \(a_{i, 0 / 1}\)
假设 \(i\)，\(j\) 对于当前的 \(mid\) 而言不能分在一起，其中 \(a_{i, 0} = a_{j, 0}\)，那么我们可以发现：

• \(i\) 上 \(a_{i, 0}\) 的课，则必须有 \(j\) 上 \(a_{j, 1}\) 的课
• \(j\) 上 \(a_{j, 0}\) 的课，则必须有 \(i\) 上 \(a_{i, 1}\) 的课

可以发现这就是一个裸的 \(\text{2-SAT}\)
所以我们每次 \(\text{check}\) 都建图跑一遍 \(\text{2-SAT}\) 就好了。
参考代码：

#include <cstring>
#include <cstdio>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
template < class T > inline T min(T a, T b) { return a < b ? a : b; }
template < class T > inline T max(T a, T b) { return a > b ? a : b; }
template < class T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while ('0' > c || c > '9') f |= c == '-', c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}
 
const int _ = 1010;
 
int tot, head[_ * 3]; struct Edge { int v, nxt; } edge[_ * _ * 4];
inline void Add_edge(int u, int v) { edge[++tot] = (Edge) { v, head[u] }, head[u] = tot; }
 
int n, t[_], a[_][2], f[_][_];
int num, dfn[_ * 3], low[_ * 3], col, co[_ * 3], top, stk[_ * 3];
 
inline void tarjan(int u) {
    dfn[u] = low[u] = ++num, stk[++top] = u;
    for (rg int i = head[u]; i; i = edge[i].nxt) {
        int v = edge[i].v;
        if (!dfn[v])
            tarjan(v), low[u] = min(low[u], low[v]);
        else
            if (!co[v]) low[u] = min(low[u], dfn[v]);
    }
    if (low[u] == dfn[u]) { ++col;
        do co[stk[top]] = col; while (stk[top--] != u);
    }
}
 
inline int id(int x, int y) { return y * (n + 1) + x; }
 
inline void init() {
    tot = num = col = top = 0;
    memset(head, 0, sizeof head);
    memset(dfn, 0, sizeof dfn);
    memset(low, 0, sizeof low);
    memset(co, 0, sizeof co);
}
 
inline bool check(int mid) {
    init();
    for (rg int i = 1; i <= n; ++i)
        for (rg int j = i + 1; j <= n; ++j) {
            if (f[i][j] <= mid) continue ;
            if (a[i][0] == a[j][0]) {
                Add_edge(id(i, a[i][0]), id(j, a[j][1]));
                Add_edge(id(j, a[j][0]), id(i, a[i][1]));
            }
            if (a[i][0] == a[j][1]) {
                Add_edge(id(i, a[i][0]), id(j, a[j][0]));
                Add_edge(id(j, a[j][1]), id(i, a[i][1]));
            }
            if (a[i][1] == a[j][0]) {
                Add_edge(id(i, a[i][1]), id(j, a[j][1]));
                Add_edge(id(j, a[j][0]), id(i, a[i][0]));
            }
            if (a[i][1] == a[j][1]) {
                Add_edge(id(i, a[i][1]), id(j, a[j][0]));
                Add_edge(id(j, a[j][1]), id(i, a[i][0]));
            }
        }
    for (rg int i = 1; i <= n; ++i)
        for (rg int k = 0; k < 3; ++k)
            if (t[i] != k && !dfn[id(i, k)]) tarjan(id(i, k));
    for (rg int i = 1; i <= n; ++i)
        if (co[id(i, a[i][0])] == co[id(i, a[i][1])]) return 0;
    return 1;
}
 
int main() {
#ifndef ONLINE_JUDGE
    file("cpp");
#endif
    read(n);
    for (rg int i = 1; i <= n; ++i) {
        read(t[i]);
        if (t[i] == 0) a[i][0] = 1, a[i][1] = 2;
        if (t[i] == 1) a[i][0] = 0, a[i][1] = 2;
        if (t[i] == 2) a[i][0] = 0, a[i][1] = 1;
        for (rg int x, j = 1; j < n; ++j) read(x), f[i][x] = j;
    }
    for (rg int i = 1; i <= n; ++i)
        for (rg int j = i + 1; j <= n; ++j) f[i][j] = max(f[i][j], f[j][i]);
    int l = 1, r = n - 1;
    while (l < r) {
        int mid = (l + r) >> 1;
        if (check(mid)) r = mid; else l = mid + 1;
    }
    printf("%d\n", l);
    return 0;
}