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;
}