Dinic 算法
/* THE PROGRAM IS MADE BY PYY */
/*----------------------------------------------------------------------------//
Copyright (c) 2011 panyanyany All rights reserved.
URL : http://poj.org/problem?id=1698
Name : 1698 Alice's Chance
Date : Wednesday, February 01, 2012
Time Stage : many hours
Result:
9763175 panyanyany
1698
Accepted 404K 16MS C++
4085B 2012-02-01 19:22:35
Test Data :
Review :
传说是最快的最大流算法，果然名不虚传啊！如果看不懂的，建议先看看这篇文章：
王欣上《浅谈基于分层思想的网络流算法》.doc
然后，一开始看的大牛的解题报告：
http://www.cnblogs.com/littlex/archive/2011/08/17/2142766.html
没有注释很伤心啊，于是我的这个注了很多释，希望以后的同学能看明白～～
//----------------------------------------------------------------------------*/
#include <cstdio>
#include <CSTRING>
using namespace std ;
#define MEM(a, v) memset (a, v, sizeof (a)) // a for address, v for value
#define max(x, y) ((x) > (y) ? (x) : (y))
#define min(x, y) ((x) < (y) ? (x) : (y))
#define INF (0x3f3f3f3f)
#define MAXN 401
#define MAXE 16000
struct EDGE {
int u, v, c, n ;
};
int n, m, eCnt ;
int map[MAXN][MAXN], dist[MAXN], vertex[MAXN], q[MAXN] ;
EDGE edge[MAXE] ;
void init ()
{
eCnt = 0 ;
MEM (vertex, -1) ;
}
void insert (int u, int v, int c)
{
edge[eCnt].u = u ;
edge[eCnt].v = v ;
edge[eCnt].c = c ;
edge[eCnt].n = vertex[u] ;
vertex[u] = eCnt++ ;
edge[eCnt].u = v ;
edge[eCnt].v = u ;
edge[eCnt].c = 0 ; // 一开始这里是赋值的 c ，结果很悲剧～～
edge[eCnt].n = vertex[v] ;
vertex[v] = eCnt++ ;
}
int dinic (int beg, int end)
{
int ans = 0 ;
while (true)
{
int head, tail, u, v, e ;
MEM(dist, -1) ;
head = tail = 0 ;
q[tail++] = beg ;
dist[beg] = 0 ;
// 广搜，构建层次图
while (head < tail)
{
v = q[head++] ;
for (e = vertex[v] ; e != -1 ; e = edge[e].n)
{
u = edge[e].u ;
int to = edge[e].v ;
int cost = edge[e].c ;
if (cost > 0 && dist[to] == -1)
{
dist[to] = dist[u] + 1 ;
q[tail++] = to ;
if (to == end)
{
head = tail ;
break ;
}
}
}
}
if (dist[end] == -1)
break ;
// v 表示增广路径的先头顶点
v = beg ;
tail = 0 ;
while (true)
{
// printf("--- tail:%d ", tail) ;
if (v == end)
{
int i, flow = INF, ebreak ;
// 寻找此路径可增加的最大流量
for (i = 0 ; i < tail ; ++i)
if (flow > edge[q[i]].c)
{
flow = edge[q[i]].c ;
ebreak = i ;
}
ans += flow ;
// 根据刚才找到的最大流，更新此路径上的所有边
for (i = 0 ; i < tail ; ++i)
{
edge[q[i]].c -= flow ; // 正向边减流
edge[q[i]^1].c += flow ; // 反向边加流
}
// 增广路径的先头顶点退至0流量的正向边的起始顶点
v = edge[q[ebreak]].u ;
tail = ebreak ;
// printf ("end --- v:%d ebreak:%d, ans:%d\n", v, ebreak, ans) ;
}
// 寻找有无可以继续增广的边
// 即，测试所有从顶点 v 起始的边中，是否有可以增广的边
// find a way from e to any vertex in "layers"
for (e = vertex[v] ; e != -1 ; e = edge[e].n)
{
// 为了避免 -1 + 1 == 0 的情况，需要测试 dist[edge[e].u] > -1
// 其实这一步貌似可以省略，因为既然能够作为增广路径的先头顶点，
// 其必然就在层次图中，因此 dist[u] 也就一定会 大于 -1
if (edge[e].c > 0 && //dist[edge[e].u] > -1 &&
dist[edge[e].u]+1 == dist[edge[e].v])
{
// printf ("dist[%d]+1 == dist[%d]: %d+1 == %d\n",
// edge[e].u, edge[e].v, dist[edge[e].u], dist[edge[e].v]) ;
break ;
}
}
// printf ("v:%d, e:%d, edge[%d]: u:%d, v:%d, c:%d, n:%d\n",
// v, e, e, edge[e].u, edge[e].v, edge[e].c, edge[e].n) ;
// system ("pause 1>>nul 2>>nul") ;
// 不能从 vertex[v] 所指向的边找到增广路
if (e == -1) // no way from current edge's next vertex
{
// 路径队列中已经没有边了
if (tail == 0) // no edges in queue
break ;
// 既然 vertex[v] 所指向的边已经无路可通了
// 那么就应该把该边的目的顶点从层次图中删除
// 一开始写成了 dist[edge[q[--tail]].u] = -1
// 结果一直死循环……本程序所有的注释代码，都是为此错误服务的……
dist[edge[q[--tail]].v] = -1 ;
// 增广路径退一条边，回到 vertex[v] 所在边的前一个顶点
v = edge[q[tail]].u ; // backward to previous vertex
// printf ("e == -1 ----- v:%d, tail:%d\n", v, tail) ;
}
else // put the edge in queue
{
// 发现一条边可用，于是加入到增广路径队列中
q[tail++] = e ;
// 将新边的目的顶点设为增广路径的先头顶点
v = edge[e].v ;
}
// puts ("") ;
}
}
return ans ;
}
int main ()
{
int i, j, k ;
int tcase, D, W, days[8], maxW, des, sum ;
while (scanf ("%d", &tcase) != EOF)
{
while (tcase--)
{
init () ;
maxW = sum = 0 ;
scanf ("%d", &n) ;
for (i = 1 ; i <= n ; ++i)
{
for (j = 1 ; j <= 7 ; ++j)
scanf ("%d", &days[j]) ;
scanf ("%d%d", &D, &W) ;
maxW = max(maxW, W) ;
sum += D ;
insert (0, i, D) ; // edges for each films
// edges from film to days
for (j = 0 ; j < W ; ++j)
for (k = 1 ; k <= 7 ; ++k)
{
if (days[k])
{
insert(i, j*7+k+n, 1) ;
}
}
}
// edges from every day to destination
des = maxW*7+n+1 ;
for (i = n + 1 ; i < des ; ++i)
insert(i, des, 1) ;
int ans = dinic (0, des) ;
puts (ans == sum ? "Yes" : "No") ;
}
}
return 0 ;
}
Edmonds_Karp 解法
/* THE PROGRAM IS MADE BY PYY */
/*----------------------------------------------------------------------------//
Copyright (c) 2011 panyanyany All rights reserved.
URL : http://poj.org/problem?id=1698
Name : 1698 Alice's Chance
Date : Saturday, January 28, 2012
Time Stage : Many hours
Result:
9749311
panyanyany
1698
Accepted
804K
860MS
C++
2637B
2012-01-28 13:52:50
Test Data :
Review :
一开始 end 是400，cnt是401，直接TLE。
//----------------------------------------------------------------------------*/
#include <cstdio>
#include <CSTRING>
#include <queue>
#include <algorithm>
#include <vector>
using namespace std ;
#define MEM(a, v) memset (a, v, sizeof (a)) // a for address, v for value
#define max(x, y) ((x) > (y) ? (x) : (y))
#define min(x, y) ((x) < (y) ? (x) : (y))
#define INF (0x3f3f3f3f)
#define MAXN 401
#define D 8
#define W 9
int n, m ;
int flow[MAXN], map[MAXN][MAXN], pre[MAXN] ;
int Mark_Point (int beg, int end, int cnt)
{
int i, t ;
queue<int> q ;
MEM (pre, -1) ;
flow[beg] = INF ;
pre[beg] = 0 ;
q.push (beg) ;
while (!q.empty ())
{
t = q.front () ;
q.pop () ;
if (t == end)
break ;
for (i = 0 ; i < cnt ; ++i)
{
if (pre[i] == -1 && map[t][i])
{
// printf ("%d-->%d ", t, i) ;
pre[i] = t ;
flow[i] = min (flow[t], map[t][i]) ;
q.push (i) ;
}
}
}
if (pre[end] == -1)
return -1 ;
return flow[end] ;
}
int Edmonds_Karp (int beg, int end, int cnt)
{
int incr, step, curr, prev ;
incr = 0 ;
while ((step = Mark_Point (beg, end, cnt)) != -1)
{
incr += step ;
curr = end ;
while (curr != beg)
{
prev = pre[curr] ;
map[prev][curr] -= step ;
map[curr][prev] += step ;
curr = prev ;
}
}
return incr ;
}
int main ()
{
int i, j, k ;
int tcase, sum, maxday ;
int w[10] ;
while (scanf ("%d", &tcase) != EOF)
{
while (tcase--)
{
scanf ("%d", &n) ;
MEM (map, 0) ;
sum = maxday = 0 ;
for (i = 1 ; i <= n ; ++i)
{
// MEM (w, 0) ;
for (j = 1 ; j <= 9 ; ++j)
scanf ("%d", &w[j]) ;
sum += w[D] ;
map[0][i] = w[D] ;
maxday = max (maxday, w[W]) ;
for (j = 0 ; j < w[W] ; ++j)
{
for (k = 1 ; k <= 7 ; ++k)
{
map[i][k+j*7+n] = w[k] ;
// printf ("%d-->%d == %d , ", i, k+j*7+20, w[k]) ;
// map[k+j*7+20][400] |= w[k] ;
// maxd = max (maxd, k+j*7+20) ;
}
}
}
maxday *= 7 ;
for (i = n + 1 ; i <= maxday + n ; ++i)
map[i][maxday+1+n] = 1 ;
// printf ("\n----%d \n", Edmonds_Karp (0, maxday+1+n, maxday+2+n)) ;
if (Edmonds_Karp (0, maxday+1+n, maxday+2+n) == sum)
puts ("Yes") ;
else
puts ("No") ;
}
}
return 0 ;
}
二分图解法
/* THE PROGRAM IS MADE BY PYY */
/*----------------------------------------------------------------------------//
Copyright (c) 2011 panyanyany All rights reserved.
URL : http://poj.org/problem?id=1698
Name : 1698 Alice's Chance
Date : Saturday, January 28, 2012
Time Stage : Many hours
Result:
9748843
panyanyany
1698
Accepted
1892K
266MS
C++
1798B
2012-01-28 10:43:31
Test Data :
Review :
网络流 dinic 算法还不会，先用二分图来做……
参考了一下解题报告：
http://blog.csdn.net/zxy_snow/article/details/6242668
//----------------------------------------------------------------------------*/
#include <cstdio>
#include <CSTRING>
#include <queue>
#include <algorithm>
#include <vector>
using namespace std ;
#define MEM(a, v) memset (a, v, sizeof (a)) // a for address, v for value
#define max(x, y) ((x) > (y) ? (x) : (y))
#define min(x, y) ((x) < (y) ? (x) : (y))
#define INF (0x3f3f3f3f)
#define MAXN 401
bool cover[MAXN] ;
int n, m, film_day ;
int map[1100][MAXN], w[10], link[MAXN] ;
int find (int cur)
{
int i ;
for (i = 1 ; i <= m ; ++i)
{
if (cover[i] == false && map[cur][i])
{
cover[i] = true ;
if (!link[i] || find (link[i]))
{
link[i] = cur ;
return true ;
}
}
}
return false ;
}
int main ()
{
int i, j, k, l ;
int tcase, sum ;
scanf ("%d", &tcase) ;
while (tcase--)
{
MEM (map, 0) ;
m = 0 ;
scanf ("%d", &n) ;
film_day = 0 ;
for (i = 1 ; i <= n ; ++i)
{
for (j = 1 ; j <= 9 ; ++j)
{
scanf ("%d", &w[j]) ;
}
for (l = film_day + 1 ; l <= film_day + w[8] ; ++l)
{
for (j = 0 ; j < w[9] ; ++j)
{
for (k = 1 ; k <= 7 ; ++k)
{
map[l][k+j*7] = w[k] ;
m = max (m, k+j*7) ;
}
}
}
film_day += w[8] ;
}
sum = 0 ;
MEM (link, 0) ;
for (i = 1 ; i <= film_day ; ++i)
{
MEM (cover, 0) ;
sum += find (i) ;
}
printf ("%s\n", sum == film_day ? "Yes" : "No") ;
}
return 0 ;
}
