组队练习赛（Regionals 2012, North America

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
",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
");
            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 += '';
    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
",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;
}