比赛的时候知道是最小支配集问题,想到了用DLX,不过没能A掉。
题意:
街霸游戏中,有n种角色,每种角色有1~2种人物,比如说Ryu有两种人:Metsu Hadoke和Metsu Shoryuken。Ryu用Metsu Hadoke可以轻易打败Chun-Li,用Metsu Shoryuke可以打败Ken。从n种角色中选出哪些角色在那种人物上,可以击败其余的任何角色的任何人物。求选取的最少人物。
解:
显然,行是代表选择,一共最多2*n行,第2*i-1行表示第i种角色选人物1,2*i行表示第i种角色选人物2。列代表约束,显然,每个角色的每种人物必须至少被覆盖一次(重复覆盖),2*n列,其次,每种角色只能选一次。增加n列,代表每种角色只能选一样,这n列必须满足每列不能多于2个1(精确覆盖)。这样,前面2*n列用重复覆盖,后面n列用精确覆盖。结束时只需判断前2*n列重复覆盖即可。
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int INF = 0x7fffffff;
const int MAX = 100;
int L[MAX*MAX],R[MAX*MAX],U[MAX*MAX],D[MAX*MAX];
int C[MAX*MAX];
int cntc[MAX];
int head;
int n,m;
int mx[MAX][MAX];
int sum[MAX];
bool vis[MAX];
struct Model
{
int k;
int beat[12][2];
};
struct Node
{
int m;
Model model[2];
};
Node node[MAX];
void EXremove(int c)
{
int i,j;
L[R[c]] = L[c];
R[L[c]] = R[c];
for(i = D[c]; i != c; i = D[i])
{
for(j = R[i]; j != i; j = R[j])
{
U[D[j]] = U[j];
D[U[j]] = D[j];
cntc[C[j]]--;
}
}
}
void EXresume(int c)
{
int i,j;
for(i = U[c]; i != c; i = U[i])
{
for(j = L[i]; j != i; j = L[j])
{
U[D[j]] = j;
D[U[j]] = j;
cntc[C[j]]++;
}
}
L[R[c]] = c;
R[L[c]] = c;
}
void remove(int c)
{
int i;
for(i = D[c]; i != c; i = D[i])
{
R[L[i]] = R[i];
L[R[i]] = L[i];
}
}
void resume(int c)
{
int i;
for(i = U[c]; i != c; i = U[i])
{
R[L[i]] = i;
L[R[i]] = i;
}
}
int h()
{
int i,j,c;
int result = 0;
memset(vis,0,sizeof(vis));
for(c = R[head]; c <= n && c != head; c = R[c])
{
if(!vis[c])
{
result++;
vis[c] = true;
for(i = D[c]; i != c; i = D[i])
{
for(j = R[i]; j != i; j = R[j])
vis[C[j]] = true;
}
}
}
return result;
}
bool dfs(int deep, int lim)
{
int i,j;
if(deep + h() > lim)
return false;
if(R[head] > n || R[head] == head)
return true;
int Min = INF,c;
for(i = R[head]; i <= n && i != head; i = R[i])
{
if(cntc[i] < Min)
{
Min = cntc[i];
c = i;
}
}
for(i = D[c]; i != c; i = D[i])
{
remove(i);
for(j = R[i]; j != i; j = R[j])
if(C[j] <= n)
remove(j);
for(j = R[i]; j != i; j = R[j])
if(C[j] > n)
EXremove(C[j]);
if(dfs(deep+1,lim))
{
//不用重新建图
for(j = L[i]; j != i; j = L[j])
if(C[j] > n)
EXresume(C[j]);
for(j = L[i]; j != i; j = L[j])
if(C[j] <= n)
resume(j);
resume(i);
return true;
}
for(j = L[i]; j != i; j = L[j])
if(C[j] > n)
EXresume(C[j]);
for(j = L[i]; j != i; j = L[j])
if(C[j] <= n)
resume(j);
resume(i);
}
return false;
}
void build()
{
int i,j;
memset(cntc,0,sizeof(cntc));
head = 0;
for(i = 0; i < m; i++)
{
R[i] = i+1;
L[i+1] = i;
}
R[m] = 0;
L[0] = m;
int first,pre,now;
//列链表
for(j = 1; j <= m; j++)
{
pre = j;
for(i = 1; i <= n; i++)
{
if(mx[i][j])
{
cntc[j]++;
now = i*m+j;
C[now] = j;
U[now] = pre;
D[pre] = now;
pre = now;
}
}
D[pre] = j;
U[j] = pre;
}
//行链表
for(i = 1; i <= n; i++)
{
pre = first = -1;
for(j = 1; j <= m; j++)
{
if(mx[i][j])
{
now = i*m+j;
if(pre == -1)
first = now;
else
{
R[pre] = now;
L[now] = pre;
}
pre = now;
}
}
if(first != -1)
{
R[pre] = first;
L[first] = pre;
}
}
}
int solve(int h)
{
int low = 1;
int high = h;
int ans = h;
while(low <= high)
{
int mid = (low+high)>>1;
if(dfs(0,mid))
{
ans = mid;
high = mid - 1;
}
else
low = mid + 1;
}
return ans;
}
int main()
{
int i,j,k;
int t,T;
scanf("%d",&T);
for(t = 1; t <= T; t++)
{
scanf("%d",&n);
m = 0;
memset(mx,0,sizeof(mx));
memset(sum,0,sizeof(sum));
for(i = 1; i <= n; i++)
{
sum[i] = m + 1;
scanf("%d",&node[i].m);
m += node[i].m;
for(j = 0; j < node[i].m; j++)
{
scanf("%d",&node[i].model[j].k);
for(k = 0; k < node[i].model[j].k; k++)
{
scanf("%d %d",&node[i].model[j].beat[k][0],&node[i].model[j].beat[k][1]);
node[i].model[j].beat[k][0]++;
}
}
}
sum[n+1] = m+1;
for(i = 1; i <= n; i++)
{
for(j = 0; j < node[i].m; j++)
{
for(k = sum[i]; k < sum[i+1]; k++)
mx[sum[i]+j][k] = 1;
for(k = 0; k < node[i].model[j].k; k++)
{
int tmpt1 = node[i].model[j].beat[k][0];
int tmpt2 = node[i].model[j].beat[k][1];
mx[sum[i]+j][sum[tmpt1]+tmpt2] = 1;
}
mx[sum[i]+j][m+i] = 1;
}
}
int tmpt = n;
n = m; //行
m += tmpt; //列 前面mi重复覆盖,后面n精确覆盖
build();
printf("Case %d: %d\n",t,solve(tmpt));
}
return 0;
}
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