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; //
const ll nmos = 0x80000000LL; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3fLL; //
const double PI=acos(-1.0); template<typename T>
inline T read(T&x){
x=;int f=;char ch=getchar();
while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
while (ch>=''&&ch<='') x=x*+ch-'',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 = ;
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=; i<=n; i++){
if(!visy[i]){
int t = wx[s] + wy[i] - mp[s][i];
if(t==){
visy[i] = true;
if(linky[i] == || dfs(linky[i])){
linkx[s] = i; linky[i] = s;
return true;
}
}
else if(t>)
{if(t<minz) minz = t;}
} }
return false;
} void km(){
for(int i=; i<=n; i++)
linkx[i] = linky[i] = ;
for(int i=; i<=n; i++){
wx[i] = -inf;
for(int j=; j<=n; j++){
wx[i] = max(wx[i], mp[i][j]);
}
wy[i] = ;
} for(int i=; i<=n; i++){
while(true){ for(int j=; j<=n; j++){
visx[j] = visy[j] = ;
} minz = inf; if(dfs(i))break;
for(int j=; j<=n; j++)
{
if(visx[j])wx[j] -= minz;
if(visy[j])wy[j] += minz;
}
}
}
}
int main(){ while(~scanf("%d", &n)){
for(int i=; i<=n; i++){
for(int j=; j<=n; j++){
scanf("%d", &mp[i][j]);
}
}
km();
ll ans = ;
for(int i=; i<=n; i++){
ans += mp[i][linkx[i]];
}
printf("%lld\n",ans);
}
return ;
}
HDU-2255