POJ 3686 The Windy's KM算法 - c++编程基础

这题的建图实在是太神了

假设某个机器处理了k个玩具,那么对于这些玩具，有两种时间，一种是真正处理的时间，一种是等待的时间，等待的时间就是之前所有处理的玩具的时间，

假设这k个玩具真正用在加工的时间分为a1,a2,a3...ak, 那么每个玩具实际的时间是加工的时间+等待时间，分别为

a1, a1+a2, a1+a2+a3.......a1+a2+...ak???

求和之后变为 a1 *k + a2 * (k - 1) + a3 * (k - 2).... + ak

这时就发现，每个玩具之间的实际时间可以分开来算? 然后求和了。

因为对每个机器，最多可以处理n个玩具，所以可以拆成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

所以我们对于每个玩具i，机器j中拆的每个点k，连接一条z[i][j]*k权值的边。

然后求最小权匹配即可

[cpp]
#include ??
#include ??
#include ??
#include ??
#include ??
#include ??
#include ??
#include ??
#include ??
#define eps 1e-5??
#define MAXN 55??
#define MAXM 55555??
#define INF 100000007??
using namespace std;?
int n, m, ny, nx;?
int w[MAXN][2555];?
int lx[MAXN], ly[2555];?
int linky[2555];?
int visx[MAXN], visy[2555];?
int slack[2555];?
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;?
}?
int 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]);?
??????????? if(d == INF) return -1;?
??????????? 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 tp = 0;?
??? for(int i = 1; i <= ny; i++)?
??????? if(linky[i] != -1) tp += w[linky[i]][i] - 5000000;?
??? return -tp;?
}?
int a[MAXN][MAXN];?
int main()?
{?
??? int T;?
??? scanf("%d", &T);?
??? while(T--)?
??? {?
??????? memset(w, 0, sizeof(w));?
??????? scanf("%d%d", &n, &m);?
??????? for(int i = 1; i <= n; i++)?
??????????? for(int j = 1; j <= m; j++)?
??????????????? scanf("%d", &a[i][j]);?
??????? for(int i = 1; i <= n; i++)?
??????????? for(int j = 1; j <= m; j++)?
??????????????? for(int k = 1; k <= n; k++)?
??????????????????? w[i][(j - 1) * n + k] = 5000000 - a[i][j] * k;?
??????? nx = n;?
??????? ny = n * m;?
??????? double ans = 1.0 * KM() / n;?
??????? printf("%f\n", ans);?
??? }?
??? return 0;?
}?

#include
#include
#include
#include
#include
#include
#include
#include
#include
#define eps 1e-5
#define MAXN 55
#define MAXM 55555
#define INF 100000007
using namespace std;
int n, m, ny, nx;
int w[MAXN][2555];
int lx[MAXN], ly[2555];
int linky[2555];
int visx[MAXN], visy[2555];
int slack[2555];
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;
}
int 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)
??????? {
???????

POJ 3686 The Windy's KM算法 (一)