Luogu 4251 [SCOI2015]小凸玩矩阵

BZOJ 4443

二分答案 + 二分图匹配

外层二分一个最小值，然后检验是否能选出$n - k + 1$个不小于当前二分出的$mid$的数。对于每一个$a_{i, j} \geq mid$，从$i$向$j + n$连一条边，然后跑二分图最大匹配即可。

菜的很，二分图匹配都写不对……

注意数组要开到双倍$n$。

时间复杂度$O(nmlogn)$。

Code：

#include <cstdio>
#include <cstring>
using namespace std;

const int N = ;
const int M = 2e5 + ;
const int inf =  << ;

int n, m, k, a[N][N], tot, head[N], mat[N];
bool vis[N];

struct Edge {
    int to, nxt;
} e[M];

inline void add(int from, int to) {
    e[++tot].to = to;
    e[tot].nxt = head[from];
    head[from] = tot;
}

inline void read(int &X) {
    X = ; char ch = ; int op = ;
    for(; ch > ''|| ch < ''; ch = getchar())
        if(ch == '-') op = -;
    for(; ch >= '' && ch <= ''; ch = getchar())
        X = (X << ) + (X << ) + ch - ;
    X *= op;
}

bool dfs(int x) {
    for(int i = head[x]; i; i = e[i].nxt) {
        int y = e[i].to;
        if(vis[y]) continue;
        vis[y] = ;
        if(!mat[y] || dfs(mat[y])) {
            mat[y] = x;
            return ;
        }
    }
    return ;
}

inline bool chk(int mid) {
    memset(head, , sizeof(head)); tot = ;
    memset(mat, , sizeof(mat));
    for(int i = ; i <= n; i++)
        for(int j = ; j <= m; j++)
            if(a[i][j] <= mid) add(i, j + n);

    int res = ;
    for(int i = ; i <= n; i++) {
        memset(vis, , sizeof(vis));
        if(dfs(i)) ++res;
    }

    return res >= n - k + ;
}

int main() {
    read(n), read(m), read(k);
    for(int i = ; i <= n; i++)
        for(int j = ; j <= m; j++)
            read(a[i][j]);

    int ln = , rn = inf, mid, res;
    for(; ln <= rn; ) {
        mid = ((ln + rn) >> );
        if(chk(mid)) rn = mid - , res = mid;
        else ln = mid + ;
    }

    printf("%d\n", res);
    return ;
}