bzoj4443

二分+二分图匹配

晚上脑子不太好使。。。

行列模型，填充数量性质，种种迹象告诉我们这是二分图，但是我觉得好像不太科学就弃了网络流。。。

二分第k大值，转化为求第n-k+1小值，二分求匹配判定即可。

#include<bits/stdc++.h>
using namespace std;
const int N = 810, inf = 1000000010;
namespace IO 
{
    const int Maxlen = N;
    char buf[Maxlen], *C = buf;
    int Len;
    inline void read_in()
    {
        Len = fread(C, 1, Maxlen, stdin);
        buf[Len] = '\0';
    }
    inline void fread(int &x) 
    {
        x = 0;
        int f = 1;
        while (*C < '0' || '9' < *C) { if(*C == '-') f = -1; ++C; }
        while ('0' <= *C && *C <= '9') x = (x << 1) + (x << 3) + *C - '0', ++C;
        x *= f;
    }
    inline void read(int &x)
    {
        x = 0;
        int f = 1; char c = getchar();
        while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
        while(c >= '0' && c <= '9') { x = (x << 1) + (x << 3) + c - '0'; c = getchar(); }
        x *= f;
    }
    inline void read(long long &x)
    {
        x = 0;
        long long f = 1; char c = getchar();
        while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
        while(c >= '0' && c <= '9') { x = (x << 1ll) + (x << 3ll) + c - '0'; c = getchar(); }
        x *= f;
    } 
} using namespace IO;
struct edge {
    int nxt, to, f;
} e[N * N * 10];
int n, m, cnt = 1, source, sink, k;
int head[N], d[N], a[N][N], iter[N];
inline void link(int u, int v, int f)
{
    e[++cnt].nxt = head[u];
    head[u] = cnt;
    e[cnt].to = v;
    e[cnt].f = f;
}
inline void insert(int u, int v, int f)
{
    link(u, v, f);
    link(v, u, 0);
}
inline bool bfs()
{
    queue<int> q;
    memset(d, -1, sizeof(d));
    d[source] = 0;
    q.push(source);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        for(int i = head[u]; i; i = e[i].nxt) if(e[i].f && d[e[i].to] == -1)
        {
            d[e[i].to] = d[u] + 1;
            q.push(e[i].to);
        }
    }
    return d[sink] != -1;
}
inline int dfs(int u, int delta)
{
    if(u == sink) return delta;
    int ret = 0;
    for(int &i = iter[u]; i && delta; i = e[i].nxt) if(e[i].f && d[e[i].to] == d[u] + 1)
    {
        int flow = dfs(e[i].to, min(e[i].f, delta));
        e[i].f -= flow;
        e[i ^ 1].f += flow;
        delta -= flow;
        ret += flow;
    }
    return ret;
}
inline int dinic()
{
    int ret = 0;
    while(bfs())
    {
        for(int i = source; i <= sink; ++i) iter[i] = head[i];
        ret += dfs(source, inf);
    }
    return ret;
}
inline bool check(int mid)
{
    memset(head, 0, sizeof(head));
    cnt = 1;
    for(int i = 1; i <= n; ++i)
    {
        insert(source, i, 1);
        for(int j = 1; j <= m; ++j) if(a[i][j] <= mid)
            insert(i, j + n, 1);
    }
    for(int i = 1; i <= m; ++i) insert(i + n, i + m + n, 1), insert(i + m + n, sink, 1);
    int ret = dinic();
    return ret >= n - k + 1;
}
int main()
{
    read(n);
    read(m);
    read(k);
    sink = 2 * m + n + 1;
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j) read(a[i][j]);
    int l = 0, r = inf + 10, ans = 0;
    while(r - l > 1)
    {
        int mid = (l + r) >> 1;
        if(check(mid)) r = ans = mid;
        else l = mid;
    }   
    printf("%d\n", ans);
    return 0;
}