hdu 5535 Cake 构造+记忆化搜索

链接：http://acm.hdu.edu.cn/showproblem.php?pid=5355

题意：给定n与m,其中1<= n <= 1e5,2 <= m <= 10;问是否存在将1~n个数分成m组，使得每组的和相等；若存在输出m行，每行表示一组的值，否则输出NO;

ps:总共T<=1000组数据，还是挺大的；

dfs也是比较巧妙，需要加个start来单调查找每组的数据，是最终全部m组全部求完了再return true，并不是每组完成就直接return，这样还可以修改直接选择的错误;

判断一组完成了只是将参数值复原为原始值；还有需要使用dp优化，设置了define来简写；

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 #define rep0(i,l,r) for(int i = (l);i < (r);i++)
  4 #define rep1(i,l,r) for(int i = (l);i <= (r);i++)
  5 #define rep_0(i,r,l) for(int i = (r);i > (l);i--)
  6 #define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
  7 #define MS0(a) memset(a,0,sizeof(a))
  8 #define MS1(a) memset(a,-1,sizeof(a))
  9 #define MSi(a) memset(a,0x3f,sizeof(a))
 10 #define inf 0x3f3f3f3f
 11 #define lson l, m, rt << 1
 12 #define rson m+1, r, rt << 1|1
 13 #define lowbit(x) (x&(-x))
 14 typedef pair<int,int> PII;
 15 #define A first
 16 #define B second
 17 #define MK make_pair
 18 typedef long long ll;
 19 typedef unsigned int uint;
 20 template<typename T>
 21 void read1(T &m)
 22 {
 23     T x=0,f=1;char ch=getchar();
 24     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 25     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
 26     m = x*f;
 27 }
 28 template<typename T>
 29 void read2(T &a,T &b){read1(a);read1(b);}
 30 template<typename T>
 31 void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
 32 template<typename T>
 33 void out(T a)
 34 {
 35     if(a>9) out(a/10);
 36     putchar(a%10+‘0‘);
 37 }
 38 int i,j,k,n,m,l,r;
 39 const int N = 1e5+7;
 40 int ans[11][N],aux[11][N],vs[44],ave,board;
 41 int dp[50][11],rec[50][11][11][50];
 42 #define def rec[board][m]
 43 bool dfs(int tot,int id,int start)
 44 {
 45     if(tot == ave){id++;tot = 0;start = 1;} // 这是一项结束了，是另一项的开始;当然了只有全部结束时，才返回true;
 46     if(id == m){
 47         rep1(i,1,board)if (!vs[i]){
 48             aux[id][++aux[id][0]] = i;
 49         }
 50         return true;
 51     }
 52     rep1(i,start,board){
 53         if(tot+i > ave) return false;
 54         if(!vs[i]){
 55             vs[i] = 1;
 56             aux[id][++aux[id][0]] = i;
 57             if(dfs(tot+i,id,i+1))  return true;
 58             vs[i] = 0,aux[id][0]--;
 59         }
 60     }
 61     return false;
 62 }
 63 bool solve(int top)
 64 {
 65     if(top >= 4*m-1){
 66         for(int i = 1,j = 0;i <= m;i++,j++){
 67             ans[i][++ans[i][0]] = top-2*m+1+j;
 68             ans[i][++ans[i][0]] = top-j;
 69         }
 70        return  solve(top-2*m);
 71     }
 72     else{
 73         ave = top*(top+1)/m/2;
 74         //printf("%d ",ave);
 75         board = top;
 76         if(dp[board][m] == 0){
 77             if(dfs(0,1,1)){
 78                 dp[board][m] = 1;
 79                 rep1(i,1,m)
 80                     rep1(j,0,aux[i][0])
 81                         def[i][j] = aux[i][j]; // define了
 82             }
 83             else dp[board][m] = -1;
 84         }
 85         if(dp[board][m] == 1) return true;
 86         return false;
 87     }
 88 }
 89 int main()
 90 {
 91     //freopen("data.txt","r",stdin);
 92     //freopen("out.txt","w",stdout);
 93     int T; read1(T);
 94     for(int kase = 1;kase <= T;kase++){
 95         MS0(vs);
 96         rep1(i,0,10) ans[i][0] = aux[i][0] = 0;
 97         read2(n,m);
 98         ll sum = 1LL*n*(n+1)/2;
 99         if(sum%m || sum/m < n || !solve(n)){
100             puts("NO");
101             continue;
102         }
103         puts("YES");
104         rep1(i,1,m){
105             printf("%d",ans[i][0]+def[i][0]);
106             rep1(j,1,ans[i][0])
107                 printf(" %d",ans[i][j]);
108             rep1(j,1,def[i][0])
109                 printf(" %d",def[i][j]);
110             puts("");
111         }
112     }
113     return 0;
114 }