POJ-3686-TheWindy's

POJ-3686-TheWindy's

http://poj.org/problem?id=3686

最大权值匹配，建图不好想，假设某个机器处理了k个玩具，时间分别为a1,a2…..,ak

那么该机器耗费的时间为a1+a1+a2+a1+a2+a3.......a1+a2+...ak  

即a1*k + a2 * (k - 1) + a3 * (k - 2).... + ak

ak玩具在某个机器上倒数第k个处理，所耗费全局的时间为ak*k

对每个机器，最多可以处理n个玩具，拆成n个点，1~n分别代表某个玩具在这个机器上倒数第几个被加工的,对于每个玩具i，机器j中拆的每个点k，连接一条map[i][j]*k权值的边

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>using namespace std;const int MAXN=5000;const int INF=0x7fffffff;int nx, ny, w[MAXN][MAXN], lx[MAXN], ly[MAXN];int fx[MAXN], fy[MAXN], matx[MAXN], maty[MAXN];int map[MAXN][MAXN];int n,m;struct node{int x;int y;}h[MAXN],man[MAXN];int path(int u){      fx[u] = 1;    for (int v = 1; v <= ny; v++)     if (lx[u] + ly[v] == w[u][v] && fy[v] < 0)    {               fy[v] = 1;            if (maty[v] < 0 || path(maty[v]))            {                    matx[u] = v;                    maty[v] = u;                    return 1;               }     }     return 0;}int km(){      int i,j,k,ret = 0;    memset(ly, 0, sizeof(ly));    for (i = 1; i <= nx; i++)    {          lx[i] = -INF;        for (j = 1; j <= ny; j++)         if (w[i][j] > lx[i]) lx[i] = w[i][j];    }    memset(matx, -1, sizeof(matx));    memset(maty, -1, sizeof(maty));    for (i = 1; i <= nx; i++)    {           memset(fx, -1, sizeof(fx));        memset(fy, -1, sizeof(fy));        if (!path(i))        {              i--;               int p = INF;            for (k = 1; k <= nx; k++){                if (fx[k] > 0)                for (j = 1; j <= ny; j++)                 if (fy[j] < 0 && lx[k] + ly[j] - w[k][j] < p)                p=lx[k]+ly[j]-w[k][j];            }            for (j = 1; j <= ny; j++)             ly[j] += fy[j]<0 ? 0 : p;            for (j = 1; j <= nx; j++)             lx[j] -= fx[j]<0 ? 0 : p;        }    }    for (i = 1; i <= ny; i++){if(maty[i]>=1)ret += w[maty[i]][i];}    return ret;} void init(){int i,j,k;scanf("%d %d",&n,&m);for(i=1;i<=n;i++)for(j=1;j<=m;j++)scanf("%d",&map[i][j]);nx=n;ny=n*m;for(i=1;i<=n;i++)for(j=1;j<=m;j++)for(k=1;k<=n;k++)    w[i][(j-1)*n+k]=-map[i][j]*k; }int main(){int t;    scanf("%d",&t);while(t--){init();printf("%.6f\n",-1.0*km()/n);}return 0;}