HDU - 2255 奔小康赚大钱 KM算法 模板题

HDU - 2255

题意： 

　　分配n所房子给n个家庭，不同家庭对一所房子所需缴纳的钱是不一样的，问你应当怎么分配房子，使得最后收到的钱最多。

思路：

　　KM算法裸题。上模板

 

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <list>
#include <cstdlib>
#include <iterator>
#include <cmath>
#include <iomanip>
#include <bitset>
#include <cctype>
using namespace std;
//#pragma GCC optimize(3)
// #pragma comment(linker, "/STACK:102400000,102400000")  //c++
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;
//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define OKC                      \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define FT(A, B, C) for (int A = B; A <= C; ++A) //用来压行
#define REP(i, j, k) for (int i = j; i < k; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFFLL;  //2147483647
const ll nmos = 0x80000000LL; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3fLL; //18
const double PI = acos(-1.0);

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}
// #define _DEBUG;         //*//
#ifdef _DEBUG
freopen("input", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
/*-----------------------show time----------------------*/
const int maxn = 309;
int mp[maxn][maxn];
int n, m;
int wx[maxn], wy[maxn];
int linkx[maxn], linky[maxn];
bool visx[maxn], visy[maxn];
int minz;
bool dfs(int s) {
    visx[s] = true;
    for (int i = 1; i <= n; i++) {
        if (!visy[i]) {
            int t = wx[s] + wy[i] - mp[s][i];
            if (t == 0) {
                visy[i] = true;
                if (linky[i] == 0 || dfs(linky[i])) {
                    linkx[s] = i;
                    linky[i] = s;
                    return true;
                }
            } else if (t > 0) {
                if (t < minz) minz = t;
            }
        }
    }
    return false;
}

void km() {
    for (int i = 1; i <= n; i++)
        linkx[i] = linky[i] = 0;
    for (int i = 1; i <= n; i++) {
        wx[i] = -inf;
        for (int j = 1; j <= n; j++) {
            wx[i] = max(wx[i], mp[i][j]);
        }
        wy[i] = 0;
    }

    for (int i = 1; i <= n; i++) {
        while (true) {
            for (int j = 1; j <= n; j++) {
                visx[j] = visy[j] = 0;
            }

            minz = inf;

            if (dfs(i)) break;
            for (int j = 1; j <= n; j++) {
                if (visx[j]) wx[j] -= minz;
                if (visy[j]) wy[j] += minz;
            }
        }
    }
}
int main() {
    while (~scanf("%d", &n)) {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                scanf("%d", &mp[i][j]);
            }
        }
        km();
        ll ans = 0;
        for (int i = 1; i <= n; i++) {
            ans += mp[i][linkx[i]];
        }
        printf("%lld\n", ans);
    }
    return 0;
}