Steady Cow Assignment
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 3062 | Accepted: 1060 |
Description
Farmer John's N (1 <= N <= 1000) cows each reside in one of B (1 <= B <= 20) barns which, of course, have limited capacity. Some cows really like their current barn, and some are not so happy.
FJ would like to rearrange the cows such that the cows are as equally happy as possible, even if that means all the cows hate their assigned barn.
Each cow gives FJ the order in which she prefers the barns. A cow's happiness with a particular assignment is her ranking of her barn. Your job is to find an assignment of cows to barns such that no barn's capacity is exceeded and the size of the range (i.e., one more than the positive difference between the the highest-ranked barn chosen and that lowest-ranked barn chosen) of barn rankings the cows give their assigned barns is as small as possible.
FJ would like to rearrange the cows such that the cows are as equally happy as possible, even if that means all the cows hate their assigned barn.
Each cow gives FJ the order in which she prefers the barns. A cow's happiness with a particular assignment is her ranking of her barn. Your job is to find an assignment of cows to barns such that no barn's capacity is exceeded and the size of the range (i.e., one more than the positive difference between the the highest-ranked barn chosen and that lowest-ranked barn chosen) of barn rankings the cows give their assigned barns is as small as possible.
Input
Line 1: Two space-separated integers, N and B
Lines 2..N+1: Each line contains B space-separated integers which are exactly 1..B sorted into some order. The first integer on line i+1 is the number of the cow i's top-choice barn, the second integer on that line is the number of the i'th cow's second-choice barn, and so on.
Line N+2: B space-separated integers, respectively the capacity of the first barn, then the capacity of the second, and so on. The sum of these numbers is guaranteed to be at least N.
Lines 2..N+1: Each line contains B space-separated integers which are exactly 1..B sorted into some order. The first integer on line i+1 is the number of the cow i's top-choice barn, the second integer on that line is the number of the i'th cow's second-choice barn, and so on.
Line N+2: B space-separated integers, respectively the capacity of the first barn, then the capacity of the second, and so on. The sum of these numbers is guaranteed to be at least N.
Output
Line 1: One integer, the size of the minumum range of barn rankings the cows give their assigned barns, including the endpoints.
Sample Input
6 4
1 2 3 4
2 3 1 4
4 2 3 1
3 1 2 4
1 3 4 2
1 4 2 3
2 1 3 2
Sample Output
2
Hint
Explanation of the sample:
Each cow can be assigned to her first or second choice: barn 1 gets cows 1 and 5, barn 2 gets cow 2, barn 3 gets cow 4, and barn 4 gets cows 3 and 6.
题意：fj想将牛重新分配到牛棚里，每只牛对所有牛棚的喜爱度不同，题目给出每只牛对牛棚的喜爱程度，最喜欢的排在最前。每个牛棚有容量限制。
Each cow can be assigned to her first or second choice: barn 1 gets cows 1 and 5, barn 2 gets cow 2, barn 3 gets cow 4, and barn 4 gets cows 3 and 6.
求：分配牛，让所有牛中最高兴的跟最不高兴的差距最小。(高兴程度：分配后，牛所在的牛棚在它心中的喜爱程度)
分析：二分枚举差距，再暴力喜爱程度的起点。就形成每只牛只能选取喜爱值从x到y的牛棚，能形成最大流，则是可行解。一直下去，直到满足最大流的最小差距。用dinic的话：从小到大枚举差距，9百多ms，用二分就2百多ms。
代码：
#include
<
stdio.h
>
#include < string .h >
#include < algorithm >
#include < iostream >
#define Min(a, b) (a) < (b) ? a : b
#define Max(a, b) (a) > (b) ? a : b
using namespace std;
const int MAXN = 2005 ;
const int MAXM = 210000 ;
const int INF = 1100000000 ;
struct Edge
{
int st, ed;
int next;
int flow;
int cap;
}edge[MAXM];
int head[MAXN], level[MAXN], que[MAXN], E;
void add( int u, int v, int w)
{
// printf("add %d %d %d\n", u, v, w);
edge[E].flow = 0 ;
edge[E].cap = w;
edge[E].st = u;
edge[E].ed = v;
edge[E].next = head[u];
head[u] = E ++ ;
edge[E].flow = 0 ;
edge[E].cap = 0 ;
edge[E].st = v;
edge[E].ed = u;
edge[E].next = head[v];
head[v] = E ++ ;
}
int dinic_bfs( int src, int dest, int ver)
{
int i, j;
for (i = 0 ; i <= ver; i ++ )
{
level[i] = - 1 ;
}
int rear = 1 ;
que[ 0 ] = src; level[src] = 0 ;
for (i = 0 ; i < rear; i ++ )
{
for (j = head[que[i]]; j != - 1 ; j = edge[j].next)
{
if (level[edge[j].ed] == - 1 && edge[j].cap > edge[j].flow)
{
level[edge[j].ed] = level[que[i]] + 1 ;
que[rear ++ ] = edge[j].ed;
}
}
}
return level[dest] >= 0 ;
}
int dinic_dfs( int src, int dest, int ver)
{
int stk[MAXN], top = 0 ;
int ret = 0 , cur, ptr, pre[MAXN], minf, i;
int del[MAXN];
for (i = 0 ; i <= ver; i ++ )
{
del[i] = 0 ;
}
stk[top ++ ] = src;
pre[src] = src;
cur = src;
while (top)
{
while (cur != dest && top)
{
for (i = head[cur]; i != - 1 ; i = edge[i].next)
{
if (level[edge[i].ed] == level[cur] + 1 && edge[i].cap > edge[i].flow && ! del[edge[i].ed])
{
stk[top ++ ] = edge[i].ed;
cur = edge[i].ed;
pre[edge[i].ed] = i;
break ;
}
}
if (i == - 1 )
{
del[cur] = 1 ;
top -- ;
if (top) cur = stk[top - 1 ];
}
}
if (cur == dest)
{
minf = INF;
while (cur != src)
{
cur = pre[cur];
if (edge[cur].cap - edge[cur].flow < minf) minf = edge[cur].cap - edge[cur].flow;
cur = edge[cur].st;
}
cur = dest;
while (cur != src)
{
cur = pre[cur];
edge[cur].flow += minf;
edge[cur ^ 1 ].flow -= minf;
if (edge[cur].cap - edge[cur].flow == 0 )
{
ptr = edge[cur].st;
}
cur = edge[cur].st;
}
while (top > 0 && stk[top - 1 ] != ptr) top -- ;
if (top) cur = stk[top - 1 ];
ret += minf;
}
}
return ret;
}
int Dinic( int src, int dest, int ver)
{
int ret = 0 , t;
while (dinic_bfs(src, dest, ver))
{
t = dinic_dfs(src, dest, ver);
if (t) ret += t;
else break ;
}
return ret;
}
int map[MAXN][ 30 ], vec[ 30 ];
void build ( int l, int r, int n, int b) // 用喜爱之在l跟r之间的边建图
{
int s = 0 , t = n + b + 1 , ver = t + 1 ;
E = 0 ;
int i, j, x, y;
for (i = 0 ; i <= ver; i ++ )
{
head[i] = - 1 ;
}
for (i = 1 ; i <= n; i ++ )
{
for (j = l; j <= r; j ++ )
{
add(i, map[i][j] + n, INF);
}
}
for (i = 1 ; i <= n; i ++ )
{
add(s, i, 1 );
}
for (i = 1 ; i <= b; i ++ )
{
add(i + n, t, vec[i]);
}
}
int main()
{
int n, b, i, j;
scanf( " %d%d " , & n, & b);
for (i = 1 ; i <= n; i ++ )
{
for (j = 1 ; j <= b; j ++ )
{
scanf( " %d " , & map[i][j]);
}
}
for (i = 1 ; i <= b; i ++ )
{
scanf( " %d " , & vec[i]);
}
int s = 0 , t = n + b + 1 , ver = t + 1 , flow, sign;
int l = 0 , r = b + 1 ;
while (l < r) // 枚举差距
{
i = (l + r) >> 1 ;
for (j = 1 , sign = 0 ; j <= b - i + 1 ; j ++ ) // 暴力喜爱值起点
{
build (j, j + i - 1 , n, b);
flow = Dinic( 0 , t, ver);
if (flow == n)
{
sign = 1 ;
break ;
}
}
if (sign)
{
r = i;
}
else
{
l = i + 1 ;
}
}
printf( " %d\n " , r);
return 0 ;
}
/*
2 2
1 2
1 2
2 1
6 4
1 2 3 4
2 3 1 4
4 2 3 1
3 1 2 4
1 3 4 2
1 4 2 3
2 1 3 2
*/
#include < string .h >
#include < algorithm >
#include < iostream >
#define Min(a, b) (a) < (b) ? a : b
#define Max(a, b) (a) > (b) ? a : b
using namespace std;
const int MAXN = 2005 ;
const int MAXM = 210000 ;
const int INF = 1100000000 ;
struct Edge
{
int st, ed;
int next;
int flow;
int cap;
}edge[MAXM];
int head[MAXN], level[MAXN], que[MAXN], E;
void add( int u, int v, int w)
{
// printf("add %d %d %d\n", u, v, w);
edge[E].flow = 0 ;
edge[E].cap = w;
edge[E].st = u;
edge[E].ed = v;
edge[E].next = head[u];
head[u] = E ++ ;
edge[E].flow = 0 ;
edge[E].cap = 0 ;
edge[E].st = v;
edge[E].ed = u;
edge[E].next = head[v];
head[v] = E ++ ;
}
int dinic_bfs( int src, int dest, int ver)
{
int i, j;
for (i = 0 ; i <= ver; i ++ )
{
level[i] = - 1 ;
}
int rear = 1 ;
que[ 0 ] = src; level[src] = 0 ;
for (i = 0 ; i < rear; i ++ )
{
for (j = head[que[i]]; j != - 1 ; j = edge[j].next)
{
if (level[edge[j].ed] == - 1 && edge[j].cap > edge[j].flow)
{
level[edge[j].ed] = level[que[i]] + 1 ;
que[rear ++ ] = edge[j].ed;
}
}
}
return level[dest] >= 0 ;
}
int dinic_dfs( int src, int dest, int ver)
{
int stk[MAXN], top = 0 ;
int ret = 0 , cur, ptr, pre[MAXN], minf, i;
int del[MAXN];
for (i = 0 ; i <= ver; i ++ )
{
del[i] = 0 ;
}
stk[top ++ ] = src;
pre[src] = src;
cur = src;
while (top)
{
while (cur != dest && top)
{
for (i = head[cur]; i != - 1 ; i = edge[i].next)
{
if (level[edge[i].ed] == level[cur] + 1 && edge[i].cap > edge[i].flow && ! del[edge[i].ed])
{
stk[top ++ ] = edge[i].ed;
cur = edge[i].ed;
pre[edge[i].ed] = i;
break ;
}
}
if (i == - 1 )
{
del[cur] = 1 ;
top -- ;
if (top) cur = stk[top - 1 ];
}
}
if (cur == dest)
{
minf = INF;
while (cur != src)
{
cur = pre[cur];
if (edge[cur].cap - edge[cur].flow < minf) minf = edge[cur].cap - edge[cur].flow;
cur = edge[cur].st;
}
cur = dest;
while (cur != src)
{
cur = pre[cur];
edge[cur].flow += minf;
edge[cur ^ 1 ].flow -= minf;
if (edge[cur].cap - edge[cur].flow == 0 )
{
ptr = edge[cur].st;
}
cur = edge[cur].st;
}
while (top > 0 && stk[top - 1 ] != ptr) top -- ;
if (top) cur = stk[top - 1 ];
ret += minf;
}
}
return ret;
}
int Dinic( int src, int dest, int ver)
{
int ret = 0 , t;
while (dinic_bfs(src, dest, ver))
{
t = dinic_dfs(src, dest, ver);
if (t) ret += t;
else break ;
}
return ret;
}
int map[MAXN][ 30 ], vec[ 30 ];
void build ( int l, int r, int n, int b) // 用喜爱之在l跟r之间的边建图
{
int s = 0 , t = n + b + 1 , ver = t + 1 ;
E = 0 ;
int i, j, x, y;
for (i = 0 ; i <= ver; i ++ )
{
head[i] = - 1 ;
}
for (i = 1 ; i <= n; i ++ )
{
for (j = l; j <= r; j ++ )
{
add(i, map[i][j] + n, INF);
}
}
for (i = 1 ; i <= n; i ++ )
{
add(s, i, 1 );
}
for (i = 1 ; i <= b; i ++ )
{
add(i + n, t, vec[i]);
}
}
int main()
{
int n, b, i, j;
scanf( " %d%d " , & n, & b);
for (i = 1 ; i <= n; i ++ )
{
for (j = 1 ; j <= b; j ++ )
{
scanf( " %d " , & map[i][j]);
}
}
for (i = 1 ; i <= b; i ++ )
{
scanf( " %d " , & vec[i]);
}
int s = 0 , t = n + b + 1 , ver = t + 1 , flow, sign;
int l = 0 , r = b + 1 ;
while (l < r) // 枚举差距
{
i = (l + r) >> 1 ;
for (j = 1 , sign = 0 ; j <= b - i + 1 ; j ++ ) // 暴力喜爱值起点
{
build (j, j + i - 1 , n, b);
flow = Dinic( 0 , t, ver);
if (flow == n)
{
sign = 1 ;
break ;
}
}
if (sign)
{
r = i;
}
else
{
l = i + 1 ;
}
}
printf( " %d\n " , r);
return 0 ;
}
/*
2 2
1 2
1 2
2 1
6 4
1 2 3 4
2 3 1 4
4 2 3 1
3 1 2 4
1 3 4 2
1 4 2 3
2 1 3 2
*/