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: sdj22251PROG: subsetLANG: 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|1using 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 1000000000using 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;}