标签：io sp for strong on bs ad amp ef
题意n选m彩票中,至少买几张彩票保证至少能中r个数.
产生两个组合s1=c(n,m)和s2=c(n,r),题意就是从s1选取最少的组合数的并集包含所有s2.
然后就是建立一个矩阵,跑一个dlx.
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#include <cstdio>
#include <cstring>
#define N 100
int n,m,r,cnt,tot;
int mustHas[N+20][9];
bool mat[N][N];
int a[10];
void dfs1(int k){
if (k>r){
cnt++;
for (int i=1;i<=r;i++){
mustHas[cnt][i] = a[i];
}
return ;
}
for (int i=a[k-1]+1;i<=n;i++){
a[k] = i;
dfs1(k+1);
a[k] = 0;
}
}
void dfs2(int k){
if (k>m){
tot++;
for (int i=1;i<=cnt;i++){
int num = 0;
for (int j=1;j<=r;j++){
for (int p=1;p<=m;p++){
if (mustHas[i][j]==a[p]){
num++;
break;
}
}
}
if (num<r) continue;
mat[tot][i] = true;
}
return ;
}
for (int i=a[k-1]+1;i<=n;i++){
a[k] = i;
dfs2(k+1);
a[k] = 0;
}
}
struct DLX{
int U[N*N],D[N*N],L[N*N],R[N*N],siz[N*N],col[N*N],vist[N*N];
int last,n,m,ans;
void init(int nn,int mm){
memset(U,0,sizeof(U));
memset(D,0,sizeof(D));
memset(L,0,sizeof(L));
memset(R,0,sizeof(R));
memset(siz,0,sizeof(siz));
memset(col,0,sizeof(col));
n = nn, m = mm;
for (int i=0;i<=m;i++){
L[i] = i - 1;
R[i] = i + 1;
U[i] = i;
D[i] = i;
}
L[0] = m ; R[m] = 0;
for (int i=m+1;i<=m+n;i++){
L[i] = i;
R[i] = i;
}
last = n + m;
ans = N;
}
void add(int x,int y){
last++;
L[last] = L[x+m];
R[last] = x+m;
L[R[last]] = last;
R[L[last]] = last;
U[last] = U[y];
D[last] = y;
U[D[last]] = last;
D[U[last]] = last;
col[last] = y;
siz[y]++;
}
void remove(int x){
for (int i=D[x];i!=x;i=D[i]){
R[L[i]] = R[i];
L[R[i]] = L[i];
}
}
void resume(int x){
for (int i=U[x];i!=x;i=U[i]){
R[L[i]] = L[R[i]] = i;
}
}
int g(){
memset(vist,0,sizeof(vist));
int res = 0;
while (1){
int k = -1;
for (int i=R[0];i;i=R[i]){
if (!vist[i]&&siz[i]>0){
k = i;
break;
}
}
if (k==-1) break;
vist[k] = true;
res++;
for (int i=D[k];i!=k;i=D[i]){
for (int j=R[i];j!=i;j=R[j]){
if (j<=m+n) continue;
vist[col[j]] = true;
}
}
}
return res;
}
void dfs(int k){
if (k+g()>=ans) return;
if (!R[0]) {
if (k<ans) ans = k;
return ;
}
int mn = N*N,x = -1;
for (int i=R[0];i;i=R[i]){
if (siz[i]<mn&&siz[i]>0){
mn = siz[i];
x = i;
}
}
if (x==-1) return;
for (int i=D[x];i!=x;i=D[i]){
if (i<=m+n) continue;
remove(i);
for (int j=R[i];j!=i;j=R[j]){
if (j<=m+n) continue;
remove(j);
}
dfs(k+1);
for (int j=L[i];j!=i;j=L[j]){
if (j<=m+n) continue;
resume(j);
}
resume(i);
}
}
}dlx;
int main(){
int t;
freopen("d:\\1.txt","a+",stdout);
int p = 0;
//for (scanf("%d",&t);t;t--){
//scanf("%d%d%d",&n,&m,&r);
for ( n=8;n<=8;n++)
for ( m=1;m<=n;m++)
for ( r=1;r<=m;r++){
memset(mustHas,sizeof(mustHas),0);
cnt = 0;
dfs1(1);
memset(mat,0,sizeof(mat));
tot = 0;
dfs2(1);
dlx.init(tot,cnt);
for (int i=1;i<=tot;i++){
for (int j=1;j<=cnt;j++){
if (mat[i][j]) dlx.add(i,j);
}
}
dlx.dfs(0);
printf("f[%d][%d][%d]=%d;\n",n,m,r,dlx.ans);
}
return 0;
}
------------------------------------------------------------------------------------
表如下:
f[1][1][1]=1;
f[2][1][1]=2;
f[2][2][1]=1;
f[2][2][2]=1;
f[3][1][1]=3;
f[3][2][1]=2;
f[3][2][2]=3;
f[3][3][1]=1;
f[3][3][2]=1;
f[3][3][3]=1;
f[4][1][1]=4;
f[4][2][1]=2;
f[4][2][2]=6;
f[4][3][1]=2;
f[4][3][2]=3;
f[4][3][3]=4;
f[4][4][1]=1;
f[4][4][2]=1;
f[4][4][3]=1;
f[4][4][4]=1;
f[5][1][1]=5;
f[5][2][1]=3;
f[5][2][2]=10;
f[5][3][1]=2;
f[5][3][2]=4;
f[5][3][3]=10;
f[5][4][1]=2;
f[5][4][2]=3;
f[5][4][3]=4;
f[5][4][4]=5;
f[5][5][1]=1;
f[5][5][2]=1;
f[5][5][3]=1;
f[5][5][4]=1;
f[5][5][5]=1;
f[6][1][1]=6;
f[6][2][1]=3;
f[6][2][2]=15;
f[6][3][1]=2;
f[6][3][2]=6;
f[6][3][3]=20;
f[6][4][1]=2;
f[6][4][2]=3;
f[6][4][3]=6;
f[6][4][4]=15;
f[6][5][1]=2;
f[6][5][2]=3;
f[6][5][3]=4;
f[6][5][4]=5;
f[6][5][5]=6;
f[6][6][1]=1;
f[6][6][2]=1;
f[6][6][3]=1;
f[6][6][4]=1;
f[6][6][5]=1;
f[6][6][6]=1;
f[7][1][1]=7;
f[7][2][1]=4;
f[7][2][2]=21;
f[7][3][1]=3;
f[7][3][2]=7;
f[7][3][3]=35;
f[7][4][1]=2;
f[7][4][2]=5;
f[7][4][3]=12;
f[7][4][4]=35;
f[7][5][1]=2;
f[7][5][2]=3;
f[7][5][3]=5;
f[7][5][4]=9;
f[7][5][5]=21;
f[7][6][1]=2;
f[7][6][2]=3;
f[7][6][3]=4;
f[7][6][4]=5;
f[7][6][5]=6;
f[7][6][6]=7;
f[7][7][1]=1;
f[7][7][2]=1;
f[7][7][3]=1;
f[7][7][4]=1;
f[7][7][5]=1;
f[7][7][6]=1;
f[7][7][7]=1;
f[8][1][1]=8;
f[8][2][1]=4;
f[8][2][2]=28;
f[8][3][1]=3;
f[8][3][2]=11;
f[8][3][3]=56;
f[8][4][1]=2;
f[8][4][2]=6;
f[8][4][3]=14;
f[8][4][4]=70;
f[8][5][1]=2;
f[8][5][2]=4;
f[8][5][3]=8;
f[8][1][1]=8;
f[8][2][1]=4;
f[8][2][2]=28;
f[8][3][1]=3;
f[8][3][2]=11;
f[8][3][3]=56;
f[8][4][1]=2;
f[8][4][2]=6;
f[8][4][3]=14;
f[8][4][4]=70;
f[8][5][1]=2;
f[8][5][2]=4;
f[8][5][3]=8;
f[8][1][1]=8;
f[8][2][1]=4;
f[8][2][2]=28;
f[8][3][1]=3;
f[8][1][1]=8;
f[8][2][1]=4;
f[8][2][2]=28;
f[8][3][1]=3;
f[8][3][2]=11;
f[8][3][3]=56;
f[8][4][1]=2;
f[8][4][2]=6;
f[8][4][3]=14;
f[8][4][4]=70;
f[8][5][1]=2;
f[8][5][2]=4;
f[8][5][3]=8;
f[8][5][4]=20;
f[8][5][5]=56;
f[8][6][1]=2;
f[8][6][2]=3;
f[8][6][3]=4;
f[8][6][4]=7;
f[8][6][5]=12;
f[8][6][6]=28;
f[8][7][1]=2;
f[8][7][2]=3;
f[8][7][3]=4;
f[8][7][4]=5;
f[8][7][5]=6;
f[8][7][6]=7;
f[8][7][7]=8;
f[8][8][1]=1;
f[8][8][2]=1;
f[8][8][3]=1;
f[8][8][4]=1;
f[8][8][5]=1;
f[8][8][6]=1;
f[8][8][7]=1;
f[8][8][8]=1;
hdu4979 A simple math problem.
标签：io sp for strong on bs ad amp ef
原文地址：http://www.cnblogs.com/nhywieza/p/4082851.html
