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 << " "; #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 ' ' #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 ",ans); } return 0; }