记忆化搜索例题记忆化搜索

1.poj 1579

题目链接：

http://poj.org/problem?id=1579

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=1e4+5;
int a,b,c;
int vis[20][20][20];
int dp[20][20][20];
int solve(int a,int b,int c)
{if(a<=0||b<=0||c<=0){return 1;}if(a>20||b>20||c>20){return solve(20,20,20);}if(a<b&&b<c){if(!vis[a][b][c]){dp[a][b][c]=solve(a,b,c-1)+solve(a,b-1,c-1)-solve(a,b-1,c);vis[a][b][c]=1;}return dp[a][b][c];}else{if(!vis[a][b][c]){dp[a][b][c]=solve(a-1,b,c)+solve(a-1,b-1,c)+solve(a-1,b,c-1)-solve(a-1,b-1,c-1);vis[a][b][c]=1;}return dp[a][b][c];}
}
int main()
{memset (vis,0,sizeof(vis));while(scanf("%d%d%d",&a,&b,&c)){if(a==-1&&b==-1&&c==-1) break;printf("w(%d, %d, %d) = %d\n",a,b,c,solve(a,b,c));}return 0;
}

2.hdu 1078 FatMouse and Cheese

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1078

dp[i][j]表示在i,j位置的最大数目.

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=105;
int n,k;
int a[maxn][maxn];
int vis[maxn][maxn];
int dp[maxn][maxn];
int pos[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
int ans;
int Max;
int Mlocx,Mlocy;
int dfs (int locx,int locy)
{if(locx==Mlocx&&locy==Mlocy){vis[Mlocx][Mlocy]=1;return 0;}if(vis[locx][locy]){return dp[locx][locy];}for (int j=1;j<=k;j++){for (int i=0;i<4;i++){int tlocx=locx+j*pos[i][0];int tlocy=locy+j*pos[i][1];if(tlocx>=1&&tlocx<=n&&tlocy>=1&&tlocy<=n){if(a[tlocx][tlocy]>a[locx][locy]){dp[locx][locy]=max(dp[locx][locy],a[tlocx][tlocy]+dfs(tlocx,tlocy));}}}}vis[locx][locy]=1;return dp[locx][locy];
}
int main()
{while(scanf("%d%d",&n,&k)){if(n==-1||k==-1) break;ans=0;Max=0;memset(dp,0,sizeof(dp));memset (vis,0,sizeof(vis));for (int i=1;i<=n;i++){for (int j=1;j<=n;j++){scanf("%d",&a[i][j]);if(Max<a[i][j]){Max=a[i][j];Mlocx=i;Mlocy=j;}}}dfs(1,1);printf("%d\n",dp[1][1]+a[1][1]);}return 0;
}

3. poj 2111 Millenium Leapcow

题目链接：

http://poj.org/problem?id=2111

思路：

设记忆化数组dp[i][j]，表示在dp[i][j]位置之后的步数，显然最大的数肯定是0，以它为终止搜索的条件。

然后进行记忆化搜索，在找到满足条件的节点的同时记下它的编号作为后继节点。

然后根据后继节点输出答案即可。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=470;
int a[maxn][maxn];
int dp[maxn][maxn];
int vis[maxn][maxn];
int Mlocx,Mlocy;
int n;
int pos[10][2]={{1,-2},{2,-1},{2,1},{1,2},{-1,2},{-2,1},{-2,-1},{-1,-2}};
int prex[maxn][maxn];
int prey[maxn][maxn];
int dfs (int locx,int locy)
{if(vis[locx][locy]){return dp[locx][locy];}if(locx==Mlocx&&locy==Mlocy){vis[locx][locy]=1;dp[locx][locy]=0;return 0;}for (int i=0;i<8;i++){int tlocx=locx+pos[i][0];int tlocy=locy+pos[i][1];if(tlocx<=0||tlocx>n||tlocy<=0||tlocy>n) continue;if(a[tlocx][tlocy]<=a[locx][locy]) continue;dfs(tlocx,tlocy);if(dp[locx][locy]<1+dp[tlocx][tlocy]){prex[locx][locy]=tlocx;prey[locx][locy]=tlocy;dp[locx][locy]=1+dp[tlocx][tlocy];}else if(dp[locx][locy]==1+dp[tlocx][tlocy]&&prex[locx][locy]!=-1&&prey[locx][locy]!=-1&&a[prex[locx][locy]][prey[locx][locy]]>a[tlocx][tlocy]){prex[locx][locy]=tlocx;prey[locx][locy]=tlocy;}}vis[locx][locy]=1;return dp[locx][locy];
}
int main()
{while(scanf("%d",&n)!=EOF){int Max=0;memset (vis,0,sizeof(vis));memset (dp,0,sizeof(dp));memset (prex,-1,sizeof(prex));memset (prey,-1,sizeof(prey));for (int i=1;i<=n;i++){for (int j=1;j<=n;j++){scanf("%d",&a[i][j]);if(Max<a[i][j]){Max=a[i][j];Mlocx=i,Mlocy=j;}}}for (int i=1;i<=n;i++){for (int j=1;j<=n;j++){dfs(i,j);}}int Maxstep=0,stx=1,sty=1;for (int i=1;i<=n;i++){for (int j=1;j<=n;j++){if(dp[i][j]>Maxstep){Maxstep=dp[i][j];stx=i,sty=j;}else if(dp[i][j]==Maxstep){if(a[i][j]<a[stx][sty]){stx=i,sty=j;}}}}printf("%d\n",dp[stx][sty]+1);int tx=stx,ty=sty;while(tx!=-1||ty!=-1){printf("%d\n",a[tx][ty]);int ttx=tx,tty=ty;tx=prex[ttx][tty],ty=prey[ttx][tty];}}return 0;
}

4.HDU 2848 Number Cutting Game

题目：

Problem Description

Two boys boyA and boyB are playing a simple game called Number Cutting Game. Now I will introduce it to you, my friend.

At the beginning of the game, a machine will generate two numbers, N(10 < N < 10 ^ 19) and K(2 <= K <= length(N)). Then boyA and boyB are take turns to play it:

When one player receives two numbers N and K, his task is to cut N into K nonempty pieces. If length(N) < K, that is to say he can not able to cut, he loses the game. Otherwise, he adds these K pieces together and get the sum SUM. Then let N = SUM and throw N and K to another player.

We assume boyA receives the two numbers from the machine firstly and they are smart enough to choose the best strategy for himself.
Now please help boyA to calculate weather he can win if he receives N and K firstly.

Here is a sample for you N = 103, K = 2.
Fisrtly, boyA receives N = 103, K = 2. He has two ways to cut like: 1|03 => 1 + 03 = 4, or 10|3 => 10 + 3 = 13.
Then, boyB receives from boyA. If boyB receives N = 4, K = 2, he loses the game. If boyB receives N = 13 and K = 2, he has only one way to cut: 1|3 = 1 + 3 = 4, then he will win the game.
So you can see boyA will win if he choose the best strategy.

Input

No more than 1000 cases. Each contains two line, first line an integer N(10 < N < 10 ^ 19), second line K(2 <= K <= length(N)).

Output

If boyA can win, print 1, otherwise 0.

Sample Input

103 2

999999999999999999 2

Sample Output

思路：

如果选择的一方没有一种办法能够组合成题目要求的情况，则判选择的一方输，否则则继续搜索下去。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
typedef unsigned long long ull;
const int maxn=470;
ull p10[30];
ull n,k;
int re;
int dfs (ull x,ull sum,int ci)
{if(ci<k-1&&x==0) return 0;if(ci==k-1){if(dfs(sum+x,0,0)) return 0;return 1;}for (int i=1;p10[i]<=x;i++){if(dfs(x/p10[i],sum+(x%p10[i]),ci+1)) return 1;}return 0;
}
int main()
{p10[0]=1;for (int i=1;i<=19;i++){p10[i]=p10[i-1]*10;}while(cin>>n>>k){re=-1;re=dfs(n,0,0);printf("%d\n",re);}return 0;
}