pku The Windy's KM最小权匹配 or 最小费用最大流
http://poj.org/problem?id=3686
题意：
给定n个玩具，有m个车间，给出每个玩具在每个车间的加工所需的时间mat[i][j]表示第i个玩具在第j个车间加工所需的时间，规顶只有第i个玩具在j车间完成时第j车间才能接受其他玩具来生产。求加工完毕所有的的n个玩具所需的最小的平均时间。
思路：
不论使用KM求最小权匹配还是使用最小费用最大流求解，建图还是最重要的。笨啊...想不到啊。。
假设有n件玩具都在第j个车间生产，他们所需要的时间分别为T1 + 2*T1 + 3*T1 +..... + n*T1; 则第c个在j上的的所需时间为c*j，
我们将每个车间分成n个点，表示第i个可能是第几个在j车间上生产。然后求一个最小权匹配，或mcmf （这里注意好好计算边与点的数量）；
KM：
View Code
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <set> #include <map> #include <string> #define CL(a,num) memset((a),(num),sizeof(a)) #define iabs(x) ((x) > 0 ? (x) : -(x)) #define Min(a,b) (a) > (b)? (b):(a) #define Max(a,b) (a) > (b)? (a):(b) #define ll long long #define inf 0x7f7f7f7f #define MOD 1073741824 #define lc l,m,rt<<1 #define rc m + 1,r,rt<<1|1 #define pi acos(-1.0) #define test puts("<------------------->") #define maxn 100007 #define M 2555 #define N 55 using namespace std; //freopen("data.in","r",stdin); int w[N][M],mat[N][N]; int lx[N],ly[M],slack[M]; bool vtx[N],vty[M]; int lk[M]; int n,m; bool dfs(int i){ vtx[i] = true; for (int j = 1; j <= m; ++j){ if (vty[j]) continue; int tmp = lx[i] + ly[j] - w[i][j]; if (tmp == 0){ vty[j] = true; if (lk[j] == -1 || dfs(lk[j])){ lk[j] = i; return true; } } else{ slack[j] = min(slack[j],tmp); } } return false; } double KM(){ int i,j; for (i = 1; i <= n; ++i){ for (j = 1,lx[i] = -inf; j <= m; ++j){ lx[i] = max(lx[i],w[i][j]); } } for (i = 1; i <= m; ++i){ lk[i] = -1; ly[i] = 0; } for (i = 1; i <= n; ++i){ for (j = 1; j <= m; ++j) slack[j] = inf; while (1){ for (j = 1; j <= n; ++j) vtx[j] = false; for (j = 1; j <= m; ++j) vty[j] = false; if (dfs(i)) break; int d = inf; for (j = 1; j <= m; ++j){ if (!vty[j]) d = min(d,slack[j]); } for (j = 1; j <= n; ++j){ if (vtx[j]) lx[j] -= d; } for (j = 1; j <= m; ++j){ if (vty[j]) ly[j] += d; else slack[j] -= d; } } } double sum = 0; for (i = 1; i <= m; ++i){ if (lk[i] > -1) sum += w[lk[i]][i]; } return (-sum); } int main(){ //freopen("data.in","r",stdin); int i,j; int k; int T; scanf("%d",&T); while (T--){ scanf("%d%d",&n,&m); for (i = 1; i <= n; ++i){ for (j = 1; j <= m; ++j){ scanf("%d",&mat[i][j]); } } CL(w,0); for (i = 1; i <= n; ++i){ for (j = 1; j <= m; ++j){ for (k = 1; k <= n; ++k){ w[i][(j - 1)*n + k] = -mat[i][j]*k; } } } m = n*m; double ans = KM()/(1.0*n); // printf("<>>>>>>>>>>>%d\n",m); printf("%.6lf\n",ans); } return 0; }
mcmf：
View Code
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <set> #include <map> #include <string> #define CL(a,num) memset((a),(num),sizeof(a)) #define iabs(x) ((x) > 0 ? (x) : -(x)) #define Min(a,b) (a) > (b)? (b):(a) #define Max(a,b) (a) > (b)? (a):(b) #define ll long long #define inf 0x7f7f7f7f #define MOD 1073741824 #define lc l,m,rt<<1 #define rc m + 1,r,rt<<1|1 #define pi acos(-1.0) #define test puts("<------------------->") #define maxn 100007 #define M 127566 #define N 53 using namespace std; //freopen("data.in","r",stdin); struct node{ int v; int w,c; int next; }g[M*2]; int head[N*N],ct; int mat[N][N]; int pre[M],dis[M]; int path[M]; bool vt[M]; int n,m; void spfa(int s,int e){ int i; for (i = s; i <= e; ++i){ dis[i] = inf; pre[i] = -1; vt[i] = false; path[i] = 0; } pre[s] = s; dis[s] = 0; vt[s] = true; queue<int>q; q.push(s); while (!q.empty()){ int u = q.front(); q.pop(); for (int i = head[u]; i != -1; i = g[i].next){ int v = g[i].v; int w = g[i].w; int ci = g[i].c; if (dis[v] > dis[u] + w && ci > 0){ dis[v] = dis[u] + w; pre[v] = u; path[v] = i; if (!vt[v]){ q.push(v); vt[v] = true; } } } vt[u] = false; } } double mcmf(int s,int e){ double ans = 0; int i; while (1){ spfa(s,e); if (pre[e] == -1) break; int MIN = inf; for (i = e; i != s; i = pre[i]){ MIN = min(MIN,g[path[i]].c); } for (i = e; i != s; i = pre[i]){ g[path[i]].c -= MIN; g[path[i]^1].c += MIN; ans += MIN*g[path[i]].w; } } return ans; } void add(int u,int v,int c,int w){ g[ct].v = v; g[ct].c = c; g[ct].w = w; g[ct].next = head[u]; head[u] = ct++; } int main(){ //freopen("data.in","r",stdin); int i,j,T; int k; scanf("%d",&T); while (T--){ CL(head,-1); ct = 0; scanf("%d%d",&n,&m); for (i = 1; i <= n; ++i){ for (j = 1; j <= m; ++j){ scanf("%d",&mat[i][j]); } } for (i = 1; i <= n; ++i){ for (j = 1; j <= m; ++j){ for (k = 1; k <= n; ++k){ int tp = j*n + k; add(i,tp,1,mat[i][j]*k); add(tp,i,0,-mat[i][j]*k); /*c[i][tp] = 1; c[tp][i] = 0; w[i][tp] = mat[i][j]*k; w[tp][i] = -mat[i][j]*k;*/ } } } int s = 0 , e = (m + 1)*n + 1; for (i = 1; i <= n; ++i){ add(s,i,1,0); add(i,s,0,0); /*c[s][i] = 1; c[i][s] = 0; w[s][i] = w[i][s] = 0;*/ } for (i = n + 1; i <= (m + 1)*n; ++i){ add(i,e,1,0); add(e,i,0,0); /*c[i][e] = 1; c[e][i] = 0; w[i][e] = w[e][i] = 0;*/ } printf("%.6lf\n",mcmf(s,e)/n); } return 0; }
