HDU 2426 Interesting Housing Problem 最小费用最大流 or KM算法
这道题貌似是08年杭州网络赛的题目， 看完题后就发现是个最优匹配问题，用KM或者最小费用最大流做， 就找了个最小费用最大流的模板，改了一下，就能过了、
建图还是比较好想的，建立一个超级源点，一个超级汇点，然后源点与学生连边，流量限制为1，费用为0，汇点与房间连边，流量限制为1， 费用为0， 然后学生与房间之间的边，流量限制是1，费用就是题目给的喜好值了， 注意，每次加边，都要加一个相应的反向边， 就是流量为0， 费用为正向边的负数。
然后题目中要求不能把学生分到不喜欢的房间里，所以碰见喜好值是负数就不管了。 另外，由于是最小费用么，所以所有喜好值都是乘以-1后加进去的，这样求出的最小值，输出的时候再乘以-1就变成最大的了。
最后判断能不能所有学生都分配到房间里， 就判断流量的大小是否等于学生个数就行了
/*
ID: sdj22251
PROG: subset
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define MAXN 100005
#define eps 1e-11
#define L(x) x<<1
#define R(x) x<<1|1
using namespace std;
int tot = 0, x, y;
class mincost
{
private:
const static int V = 2001; //注意点的个数， 应该为学生+房间+2， 所以至少要开到1002
const static int E = 2000001;
const static int INF = -1u >> 1;
struct Edge
{
int v, cap, cost;
Edge *next;
} pool[E], *g[V], *pp, *pree[V];
int T, S, dis[V], pre[V];
int n, m, flow, cirq[V];
void SPFA();
inline void addedge(int i, int j, int cap, int cost);
public:
bool initialize(int x, int y, int z);
void mincost_maxflow();
};
void mincost::mincost_maxflow()
{
while (true)
{
SPFA();
if (dis[T] == INF)
break;
int minn = INF;
for (int i = T; i != S; i = pre[i])
minn = min(minn, pree[i]->cap);
for (int i = T; i != S; i = pre[i])
{
pree[i]->cap -= minn;
pool[(pree[i] - pool)^0x1].cap += minn;
flow += minn * pree[i]->cost;
}
tot += minn; //流量计算
}
if(tot != x) flow = 1;
printf("%d\n", -flow);
}
void mincost::SPFA()
{
bool vst[V] = {false};
int tail = 0, u;
fill(dis,dis + n,0x7fffffff);
cirq[0] = S;
vst[S] = 1;
dis[S] = 0;
for (int i = 0; i <= tail; i++)
{
int v = cirq[i % n];
for (Edge *i = g[v]; i != NULL; i = i->next)
{
if (!i->cap)
continue;
u = i->v;
if (i->cost + dis[v] < dis[u])
{
dis[u] = i->cost + dis[v];
pree[u] = i;
pre[u] = v;
if (!vst[u])
{
tail++;
cirq[tail % n] = u;
vst[u] = true;
}
}
}
vst[v] = false;
}
}
void mincost::addedge(int i, int j, int cap, int cost)
{
pp->cap = cap;
pp->v = j;
pp->cost = cost;
pp->next = g[i];
g[i] = pp++;
}
bool mincost::initialize(int x, int y, int z)
{
memset(g, 0, sizeof (g));
pp = pool;
n = x + y + 2; //n即为顶点的个数 学生数+房间数+一个源点+一个汇点
m = z;
S = 0;
T = x + y + 1;
int v, u, f, c;
for (int i = 0; i < m; i++)
{
scanf("%d %d %d", &v, &u, &c);
v++;
u++;
if(c >= 0)
{
addedge(v, u + x, 1, -c);
addedge(u + x, v, 0, c);
}
}
for(int i = 1; i <= x; i++)
{
addedge(0, i, 1, 0);
addedge(i, 0, 0, 0);
}
for(int i = 1; i <= y; i++)
{
addedge(x + i, T, 1, 0);
addedge(T, x + i, 0, 0);
}
flow = 0;
return true;
}
mincost g;
int main()
{
//freopen("d:/data.in","r",stdin);
//freopen("d:/data.out","w",stdout);
int z, cas = 0;
while(scanf("%d%d%d", &x, &y, &z) != EOF)
{
g.initialize(x, y, z);
printf("Case %d: ", ++cas);
tot = 0;
g.mincost_maxflow();
}
return 0;
}
接下来是KM算法的 最后判断是否完备匹配就行了
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#define MAXN 505
#define MAXM 555555
#define INF 1000000000
using namespace std;
int n, m, ny, nx;
int w[MAXN][MAXN];
int lx[MAXN], ly[MAXN];
int linky[MAXN];
int visx[MAXN], visy[MAXN];
int slack[MAXN];
bool find(int x)
{
visx[x] = 1;
for(int y = 1; y <= ny; y++)
{
if(visy[y]) continue;
int t = lx[x] + ly[y] - w[x][y];
if(t == 0)
{
visy[y] = 1;
if(linky[y] == -1 || find(linky[y]))
{
linky[y] = x;
return true;
}
}
else if(slack[y] > t) slack[y] = t;
}
return false;
}
void KM()
{
memset(linky, -1, sizeof(linky));
for(int i = 1; i <= nx; i++) lx[i] = -INF;
memset(ly, 0, sizeof(ly));
for(int i = 1; i <= nx; i++)
for(int j = 1; j <= ny; j++)
if(w[i][j] > lx[i]) lx[i] = w[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(find(x)) break;
int d = INF;
for(int i = 1; i <= ny; i++)
if(!visy[i]) d = min(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 main()
{
int x, y, z, e, cas = 0;
while(scanf("%d%d%d", &n, &m, &e) != EOF)
{
nx = n, ny = m;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
w[i][j] = -INF;
while(e--)
{
scanf("%d%d%d", &x, &y, &z);
if(z >= 0) w[x + 1][y + 1] = z;
}
printf("Case %d: ", ++cas);
KM();
int cnt = 0, ans = 0;
for(int i = 1; i <= ny; i++)
if(linky[i] != -1 && w[linky[i]][i] != -INF)
ans += w[linky[i]][i], cnt++;
if(cnt != nx) ans = -1;
printf("%d\n", ans);
}
return 0;
}