題意：最裸的帶權二分圖匹配。

思路：KM演算法，複雜度為O(n^3)，比費用流的效率要快一些，所以有時候必須用KM演算法。

但是KM演算法的適用範圍比較小，要學會去分辨.

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define fuck(x) cout<<"["<<x<<"]";
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;

const int MX = 3e2 + 5;
/*  KM演算法
 *   複雜度O(nx*nx*ny)
 *  求最大權匹配
 *   若求最小權匹配，可將權值取相反數，結果取相反數
 *  點的編號從1開始
 */
const int INF = 0x3f3f3f3f;
int nx, ny; //兩邊的點數
int G[MX][MX];//二分圖描述
int linker[MX], lx[MX], ly[MX]; //y中各點匹配狀態，x,y中的點標號
int slack[MX];
bool visx[MX], visy[MX];

bool DFS(int x) {
    visx[x] = 1;
    for(int y = 1; y <= ny; y++) {
        if(visy[y]) continue;
        int tmp = lx[x] + ly[y] - G[x][y];
        if(tmp == 0) {
            visy[y] = 1;
            if(linker[y] == -1 || DFS(linker[y])) {
                linker[y] = x;
                return 1;
            }
        } else if(slack[y] > tmp) {
            slack[y] = tmp;
        }
    }
    return 0;
}
int KM() {
    memset(linker, -1, sizeof(linker));
    memset(ly, 0, sizeof(ly));
    for(int i = 1; i <= nx; i++) {
        lx[i] = -INF;
        for(int j = 1; j <= ny; j++) {
            if(G[i][j] > lx[i]) lx[i] = G[i][j];
        }
    }
    for(int x = 1; x <= nx; x++) {
        for(int i = 1; i <= ny; i++) slack[i] = INF;
        while(true) {
            memset(visx, 0, sizeof(visx));
            memset(visy, 0, sizeof(visy));
            if(DFS(x)) break;
            int d = INF;
            for(int i = 1; i <= ny; i++) {
                if(!visy[i] && d > slack[i]) d = slack[i];
            }
            for(int i = 1; i <= nx; i++) {
                if(visx[i]) lx[i] -= d;
            }
            for(int i = 1; i <= ny; i++) {
                if(visy[i]) ly[i] += d;
                else slack[i] -= d;
            }
        }
    }
    int res = 0;
    for(int i = 1; i <= ny; i++) {
        if(linker[i] != -1) res += G[linker[i]][i];
    }
    return res;
}
int main() {
    int n; //FIN;
    while(~scanf("%d", &n)) {
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                scanf("%d", &G[i][j]);
            }
        }
        nx = ny = n;
        printf("%d\n", KM());
    }
    return 0;
}