【POJ2385】Apple Catching(简单DP)

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记录一个菜逼的成长。。

题目大意：两棵树，有T秒钟，每秒钟都会下落一个苹果，接下来T行表示下落苹果的树的编号。最多有W次交换次数，初始在1树。
dp[i][j] = max(dp[i-1][j],dp[i-1][j-1]);
还要判断现在所在的树是否有苹果落下。
还要加上(j%2) + 1 == tree[i]; 0次是在1树，所以如果交换次数是奇数则在第二棵树，判断i秒钟苹果是否在第二棵树落下
PS:DP的题还是要记一下。。话说刚开始没看懂什么意思。。

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <cstdlib>#include <vector>#include <set>#include <map>#include <queue>#include <stack>#include <list>#include <deque>#include <cctype>#include <bitset>#include <cmath>using namespace std;#define ALL(v) (v).begin(),(v).end()#define cl(a) memset(a,0,sizeof(a))#define bp __builtin_popcount#define pb push_back#define fin freopen("D://in.txt","r",stdin)#define fout freopen("D://out.txt","w",stdout)#define lson t<<1,l,mid#define rson t<<1|1,mid+1,r#define seglen (node[t].r-node[t].l+1)#define pi 3.1415926#define e  2.718281828459typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> PII;typedef pair<LL,LL> PLL;typedef vector<PII> VPII;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;template <typename T>inline void read(T &x){    T ans=0;    char last=' ',ch=getchar();    while(ch<'0' || ch>'9')last=ch,ch=getchar();    while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();    if(last=='-')ans=-ans;    x = ans;}inline bool DBread(double &num){    char in;double Dec=0.1;    bool IsN=false,IsD=false;    in=getchar();    if(in==EOF) return false;    while(in!='-'&&in!='.'&&(in<'0'||in>'9'))        in=getchar();    if(in=='-'){IsN=true;num=0;}    else if(in=='.'){IsD=true;num=0;}    else num=in-'0';    if(!IsD){        while(in=getchar(),in>='0'&&in<='9'){            num*=10;num+=in-'0';}    }    if(in!='.'){        if(IsN) num=-num;            return true;    }else{        while(in=getchar(),in>='0'&&in<='9'){                num+=Dec*(in-'0');Dec*=0.1;        }    }    if(IsN) num=-num;    return true;}template <typename T>inline void write(T a) {    if(a < 0) { putchar('-'); a = -a; }    if(a >= 10) write(a / 10);    putchar(a % 10 + '0');}/******************head***********************/const int maxn = 1000 + 10;int dp[maxn][35],tree[maxn];int main(){    int T,W;    while(~scanf("%d%d",&T,&W)){        cl(dp);        for( int i = 1; i <= T; i++ ){            scanf("%d",tree+i);        }        for( int i = 1; i <= T; i++ ){            dp[i][0] = dp[i-1][0] + (tree[i] == 1);//交换0次是在1树，判断此时是否有苹果落下。            for( int j = 1; j <= W; j++ ){                dp[i][j] = max(dp[i-1][j] , dp[i-1][j-1]) + ((j % 2) + 1 == tree[i]);            }        }        int ans = 0;        //不一定是交换次数最大 ，收到的苹果最多。        for( int i = 1; i <= W; i++ ){            ans = max(dp[T][i],ans);        }        printf("%d\n",ans);    }    return 0;}

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