组队练习赛（Regionals 2012, North America - East Central NA）
A.Babs' Box Boutique
给定n个盒子，每个盒子都有长宽高（任意两个盒子长宽高不完全相同），现在选盒子的任意两面，要求x1 <= x2 && y1 <= y2,问最多能选多少盒子满足这需求。
直接dfs暴搞................
#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <map>
# define INF 0x7FFFFFFF
using namespace std;
int vis[11];
struct node {
int a[3];
}p[11];
int ans,n;
int dx[] = {0,0,1};
int dy[] = {1,2,2};
//bool cmp(int a,int b) {
// return a > b;
//}
void dfs(int step,int x,int y) {
if(step > ans) ans = step;
if(step == n) return ;
for(int i=0; i<n; i++) {
if(vis[i] == 1) continue;
for(int j=0; j<3; j++) {
int xx = p[i].a[dx[j]];
int yy = p[i].a[dy[j]];
if(xx >= x && yy >= y ) {
vis[i] = 1;
dfs(step + 1,xx,yy);
vis[i] = 0;
}
}
}
}
int main() {
int casee = 1;
while(scanf("%d",&n) && n) {
if(n == 0) break;
int tmp[3];
for(int i=0; i<n; i++) {
scanf("%d%d%d",&tmp[0],&tmp[1],&tmp[2]);
sort(tmp,tmp+3);
p[i].a[0] = tmp[0];
p[i].a[1] = tmp[1];
p[i].a[2] = tmp[2];
}
ans = 0;
memset(vis,0,sizeof(vis));
dfs(0,0,0);
printf("Case %d: %d\n",casee++,ans);
}
return 0;
}
C.Hexagon Perplexagon
题意就不写了：戳这
dfs暴搞，每一层传入三个参数，层数， 每一块需要匹配的前驱点，和最后一块需要匹配的点..........
#include <iostream>
#include <stdio.h>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
int map[8][8];
int ans[8];
int save[8];
int vis[8];
int n,ok;
void change(int x,int pos) {
int i;
for(i=0; i<6; i++) {
if(map[x][i] == pos) break;
}
pos = i;
int cnt = 0;
for(int i=0; i<6; i++) {
save[cnt++] = map[x][pos];
pos ++;
if(pos >= 6) pos -= 6;
}
}
int getpre(int pos,int t) {
t --;
if(t < 0) t = 5;
return map[pos][t];
}
int getsuc(int pos,int t) {
t ++;
if(t > 5) t = 0;
return map[pos][t];
}
void dfs(int step,int p1,int p2) {
if(step == 7) return ;
if(ok == 1) return ;
for(int i=0; i<7; i++) {
if(vis[i] == 1) continue;
int j;
for(j=0; j<6; j++) {
if(map[i][j] == save[step-1]) break;
}
int tmp1 = getpre(i,j);
int tmp2 = getsuc(i,j);
if(step == 1) {
ans[step] = i;
vis[i] = 1;
dfs(step + 1,tmp1,tmp2);
vis[i] = 0;
}
if(step > 1 && step < 6) {
if(tmp2 == p1 ) {
ans[step] = i;
vis[i] = 1;
dfs(step+1,tmp1,p2);
vis[i] = 0;
}
}
if(step == 6) {
if(tmp2 == p1 && tmp1 == p2 ) {
ans[step] = i;
ok = 1;
}
}
if(ok == 1) return ;
}
}
int main() {
int T;
cin >> T;
int casee = 1;
while(T --) {
for(int i=0; i<7; i++) {
for(int j=0; j<6; j++) {
scanf("%d",&map[i][j]);
}
}
ok = 0;
for(int i=0; i<7; i++) {
memset(vis,0,sizeof(vis));
vis[i] = 1;
change(i,1);
ans[0] = i;
dfs(1,0,0);
if(ok) break;
}
printf("Case %d: ",casee++);
if(ok == 0) {
printf("No solution\n");
continue;
}
printf("%d",ans[0]);
for(int i=1; i<7; i++) {
printf(" %d",ans[i]);
}
puts("");
}
return 0;
}
D.I've Got Your Back(gammon)
暴搞，每一种排列分别保存它的各个点的情况，并且用map保存每种排列的顺序下标
#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <cmath>
#include <map>
using namespace std;
struct node
{
int pos[10];
} p[20000];
map <string,int>mm;
int cnt ;
string str = "";
string str2 = "";
string change(int x)
{
str = "";
str2 = "";
if(x == 0)
{
str += '0';
return str;
}
int cn = 0;
while(x)
{
int t = x % 10;
str += t + '0';
x = x / 10;
}
int len = str.length();
for(int i=len-1; i>=0; i--) str2 += str[i];
str2 += '\0';
return str2;
}
void init()
{
cnt = 0;
for(int i=0; i<=15; i++)
{
for(int j=0; j<=15; j++)
{
for(int k=0; k<=15; k++)
{
for(int l=0; l<=15; l++)
{
for(int x=0; x<=15; x++)
{
for(int y=0; y<=15; y++)
{
int t = i + j + k + l + x + y;
if(t > 15) break;
if(t == 15)
{
string tmp;
p[cnt].pos[0] = i;
p[cnt].pos[1] = j;
p[cnt].pos[2] = k;
p[cnt].pos[3] = l;
p[cnt].pos[4] = x;
p[cnt].pos[5] = y;
tmp += change(i);
tmp += ' ';
tmp += change(j);
tmp += ' ';
tmp += change(k);
tmp += ' ';
tmp += change(l);
tmp += ' ';
tmp += change(x);
tmp += ' ';
tmp += change(y);
mm[tmp] = cnt;
cnt ++;
}
}
}
}
}
}
}
}
int n;
int main()
{
init();
char c;
int a,b,d,e,f,g;
int casee = 1;
while(cin >> c)
{
if(c == 'e') break;
printf("Case %d: ",casee++);
if(c == 'm')
{
scanf("%d%d%d%d%d%d",&a,&b,&d,&e,&f,&g);
string tmp;
tmp += change(a);
tmp += ' ';
tmp += change(b);
tmp += ' ';
tmp += change(d);
tmp += ' ';
tmp += change(e);
tmp += ' ';
tmp += change(f);
tmp += ' ';
tmp += change(g);
//cout << tmp << endl;
printf("%d\n",mm[tmp]);
}
if(c == 'u')
{
scanf("%d",&a);
printf("%d",p[a].pos[0]);
for(int i=1; i<6; i++)
{
printf(" %d",p[a].pos[i]);
}
puts("");
}
}
return 0;
}