1099. Work Scheduling
Time Limit: 0.5 second
Memory Limit: 16 MB
There is certain amount of night guards that are available to protect the local junkyard from possible junk robberies. These guards need to scheduled in pairs, so that each pair guards at different night. The junkyard CEO ordered you to write a program which given the guards characteristics determines the maximum amount of scheduled guards (the rest will be fired). Please note that each guard can be scheduled with only one of his colleagues and no guard can work alone.
Input
The first line of the input contains one number N ≤ 222 which is the amount of night guards. Unlimited number of lines consisting of unordered pairs (i, j) follow, each such pair means that guard #i and guard #j can work together, because it is possible to find uniforms that suit both of them (The junkyard uses different parts of uniforms for different guards i.e. helmets, pants, jackets. It is impossible to put small helmet on a guard with a big head or big shoes on guard with small feet). The input ends with Eof.
Output
You should output one possible optimal assignment. On the first line of the output write the even number C, the amount of scheduled guards. Then output C/2 lines, each containing 2 integers (i, j) that denote that i and j will work together.
Sample
| input | output |
|---|
3
1 2
2 3
1 3
| 2
1 2
|
Problem Author: Jivko Ganev
模板題.....一般圖匹配主要看模板，剩下的就是自己YY建圖了.....
模板：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#define MAXN 256
using namespace std;
struct Graph
{
bool mat[MAXN + 1][MAXN + 1];
int n;
bool inque[MAXN + 1];
int que[MAXN], head, tail;
int match[MAXN + 1], father[MAXN + 1], base[MAXN + 1];
void init(int _n)
{
n = _n;
for (int i = 1; i <= n; ++i)
{
match[i] = 0;
for (int j = 1; j <= n; ++j)
mat[i][j] = false;
}
}
int pop()
{
return que[head++];
}
void push(int x)
{
que[tail++] = x;
inque[x] = true;
}
void add_edge(int a, int b)
{
mat[a][b] = mat[b][a] = true;
}
int inpath[MAXN + 1];
static int pcnt;
int find_ancestor(int u, int v)
{
++pcnt;
while (u)
{
u = base[u];
inpath[u] = pcnt;
u = father[match[u]];
}
while (true)
{
v = base[v];
if (inpath[v] == pcnt)
return v;
v = father[match[v]];
}
}
int inblossom[MAXN + 1];
static int bcnt;
void reset(int u, int anc)
{
while (u != anc)
{
int v = match[u];
inblossom[base[v]] = bcnt;
inblossom[base[u]] = bcnt;
v = father[v];
if (base[v] != anc) father[v] = match[u];
u = v;
}
}
void contract(int u, int v)
{
int anc = find_ancestor(u, v);
++bcnt;
reset(u, anc);
reset(v, anc);
if (base[u] != anc) father[u] = v;
if (base[v] != anc) father[v] = u;
for (int i = 1; i <= n; ++i)
if (inblossom[base[i]] == bcnt)
{
base[i] = anc;
if (!inque[i]) push(i);
}
}
int find_augment(int start)
{
for (int i = 1; i <= n; ++i)
{
father[i] = 0;
inque[i] = false;
base[i] = i;
}
head = 0, tail = 0, push(start);
while (head < tail)
{
int u = pop();
for (int v = 1; v <= n; ++v)
if (mat[u][v] && base[v] != base[u] && match[v] != u)
{
if (v == start || (match[v] && father[match[v]]))
contract(u, v);
else
{
if (father[v] == 0)
{
if (match[v])
{
push(match[v]);
father[v] = u;
}
else
{
father[v] = u;
return v;
}
}
}
}
}
return 0;
}
void augment(int finish)
{
int u = finish, v, w;
while (u)
{
v = father[u];
w = match[v];
match[u] = v;
match[v] = u;
u = w;
}
}
int graph_max_match()
{
int ans = 0;
for (int i = 1; i <= n; ++i)
if (match[i] == 0)
{
int finish = find_augment(i);
if (finish)
{
augment(finish);
ans += 2;
}
}
return ans;
}
} g;
int Graph :: bcnt = 0, Graph :: pcnt = 0;
int main()
{
int n;
scanf("%d", &n);
g.init(n);
int a, b;
while (scanf("%d%d", &a, &b) != EOF)
g.add_edge(a, b);
printf("%d\n", g.graph_max_match());
for (int i = 1; i <= n; ++i)
if (g.match[i])
{
printf("%d %d\n", i, g.match[i]);
g.match[g.match[i]] = 0;
}
return 0;
}
posted on 2011-10-15 22:16
LLawliet 閱讀(305)
評論(0) 編輯 收藏 引用 所屬分類:
圖論