hdu3498 whosyourdaddy--可重复覆盖舞蹈链
2016-09-27 18:06
316 查看
原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=3498
题意:n个怪兽,m对怪兽的关系,表示怪兽a和b之间连线,当把一个怪兽杀死,和其直接相连的怪兽都死,求最少杀多少个可以使所有怪兽死亡。
#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<vector>
#include<cstring>
#include<queue>
#include<stack>
#include<algorithm>
#include<cmath>
#define INF 99999999
#define eps 0.0001
using namespace std;
const int MAXC = 55 + 10;
const int MAXR = 55 + 10;
const int MAXN = MAXC*MAXR + 100;
int n, m;
int cnt;
int ans;
int p[60][60];
int vis[60];
int U[MAXN], D[MAXN], L[MAXN], R[MAXN];
int C[MAXN];
int H[MAXR];
int S[MAXC];
void init(int n)
{
for (int i = 0; i <= n; i++)
{
U[i] = D[i] = i;
L[i] = i - 1;
R[i] = i + 1;
S[i] = 0;
}
L[0] = n;
R
= 0;
cnt = n + 1;
memset(H, 0, sizeof(H));
}
void link(int r, int c)
{
S[c]++;
C[cnt] = c;
U[cnt] = U[c];
D[U[c]] = cnt;
D[cnt] = c;
U[c] = cnt;
if (H[r] == 0)
H[r] = L[cnt] = R[cnt] = cnt;
else
{
L[cnt] = L[H[r]];
R[L[H[r]]] = cnt;
R[cnt] = H[r];
L[H[r]] = cnt;
}
cnt++;
}
void remove(int cnt_)
{
for (int i = D[cnt_]; i != cnt_; i = D[i])
{
L[R[i]] = L[i];
R[L[i]] = R[i];
}
}
void resume(int cnt_)
{
for (int i = D[cnt_]; i != cnt_; i = D[i])
L[R[i]] = R[L[i]] = i;
}
int e()
{
int num = 0;
memset(vis, 0, sizeof(vis));
for (int i = R[0]; i != 0; i = R[i])
{
if (!vis[i])
{
num++;
for (int j = D[i]; j != i; j = D[j])
for (int k = R[j]; k != j; k = R[k])
vis[C[k]] = 1;
}
}
return num;
}
void dance(int k)
{
if (k + e() >= ans)
return;
if (R[0] == 0)
{
if (k < ans)
ans = k;
return;
}
int c;
int minn = INF;
for (int i = R[0]; i != 0; i = R[i])
{
if (minn > S[i])
{
minn = S[i];
c = i;
}
}
for (int i = D[c]; i != c; i = D[i])
{
remove(i);
for (int j = R[i]; j != i; j = R[j])
remove(j);
dance(k + 1);
for (int j = R[i]; j != i; j = R[j])
resume(j);
resume(i);
}
}
int main()
{
int a, b;
while (~scanf("%d%d", &n, &m))
{
memset(p, 0, sizeof(p));
init(n);
while (m--)
{
scanf("%d%d", &a, &b);
p[a][b] = p[b][a] = 1;
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
if (p[i][j] || i == j)
link(i, j);
ans = 100;
dance(0);
printf("%d\n", ans);
}
return 0;
}
题意:n个怪兽,m对怪兽的关系,表示怪兽a和b之间连线,当把一个怪兽杀死,和其直接相连的怪兽都死,求最少杀多少个可以使所有怪兽死亡。
#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<vector>
#include<cstring>
#include<queue>
#include<stack>
#include<algorithm>
#include<cmath>
#define INF 99999999
#define eps 0.0001
using namespace std;
const int MAXC = 55 + 10;
const int MAXR = 55 + 10;
const int MAXN = MAXC*MAXR + 100;
int n, m;
int cnt;
int ans;
int p[60][60];
int vis[60];
int U[MAXN], D[MAXN], L[MAXN], R[MAXN];
int C[MAXN];
int H[MAXR];
int S[MAXC];
void init(int n)
{
for (int i = 0; i <= n; i++)
{
U[i] = D[i] = i;
L[i] = i - 1;
R[i] = i + 1;
S[i] = 0;
}
L[0] = n;
R
= 0;
cnt = n + 1;
memset(H, 0, sizeof(H));
}
void link(int r, int c)
{
S[c]++;
C[cnt] = c;
U[cnt] = U[c];
D[U[c]] = cnt;
D[cnt] = c;
U[c] = cnt;
if (H[r] == 0)
H[r] = L[cnt] = R[cnt] = cnt;
else
{
L[cnt] = L[H[r]];
R[L[H[r]]] = cnt;
R[cnt] = H[r];
L[H[r]] = cnt;
}
cnt++;
}
void remove(int cnt_)
{
for (int i = D[cnt_]; i != cnt_; i = D[i])
{
L[R[i]] = L[i];
R[L[i]] = R[i];
}
}
void resume(int cnt_)
{
for (int i = D[cnt_]; i != cnt_; i = D[i])
L[R[i]] = R[L[i]] = i;
}
int e()
{
int num = 0;
memset(vis, 0, sizeof(vis));
for (int i = R[0]; i != 0; i = R[i])
{
if (!vis[i])
{
num++;
for (int j = D[i]; j != i; j = D[j])
for (int k = R[j]; k != j; k = R[k])
vis[C[k]] = 1;
}
}
return num;
}
void dance(int k)
{
if (k + e() >= ans)
return;
if (R[0] == 0)
{
if (k < ans)
ans = k;
return;
}
int c;
int minn = INF;
for (int i = R[0]; i != 0; i = R[i])
{
if (minn > S[i])
{
minn = S[i];
c = i;
}
}
for (int i = D[c]; i != c; i = D[i])
{
remove(i);
for (int j = R[i]; j != i; j = R[j])
remove(j);
dance(k + 1);
for (int j = R[i]; j != i; j = R[j])
resume(j);
resume(i);
}
}
int main()
{
int a, b;
while (~scanf("%d%d", &n, &m))
{
memset(p, 0, sizeof(p));
init(n);
while (m--)
{
scanf("%d%d", &a, &b);
p[a][b] = p[b][a] = 1;
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
if (p[i][j] || i == j)
link(i, j);
ans = 100;
dance(0);
printf("%d\n", ans);
}
return 0;
}
相关文章推荐
- HDU 3498 whosyourdaddy (可重复覆盖舞蹈链)
- hdu 3498 whosyourdaddy (重复覆盖,DLX+迭代加深A*)
- hdu - 3498 - whosyourdaddy(重复覆盖DLX)
- HDU 3498 whosyourdaddy DLX重复覆盖
- HDU 3498 whosyourdaddy 重复覆盖 DLX+A*
- Hdu3498-whosyourdaddy(精确覆盖模板题)
- hdu 3498 whosyourdaddy(重复覆盖+估价函数剪枝)
- [DLX重复覆盖] hdu 3498 whosyourdaddy
- 【HDU】3498 whosyourdaddy 重复覆盖入门题
- HDU 3498 whosyourdaddy【Dancing Links重复覆盖】
- HDU 4398 whosyourdaddy 精确覆盖,允许重复覆盖
- 【DLX算法】hdu3498 whosyourdaddy
- HDU 3498 whosyourdaddy(DLX+A*||多重覆盖)
- 【DLX】 hdu3498 whosyourdaddy
- HDU3498-whosyourdaddy
- DLX 舞蹈链 精确覆盖 与 重复覆盖
- HDU 2828 舞蹈链可重复覆盖
- HDU 3498 whosyourdaddy
- 【HDU】3498 whosyourdaddy
- HDU 3663 舞蹈链之不可重复覆盖