LightOJ 1004 - Monkey Banana Problem

题意：简单的数塔，注意分成两部分，转移时两部分有不同。

AC代码：

////  Created by  CQU_CST_WuErli//  Copyright (c) 2015 CQU_CST_WuErli. All rights reserved.//// #include<bits/stdc++.h>#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <cctype>#include <cmath>#include <string>#include <vector>#include <list>#include <map>#include <queue>#include <stack>#include <set>#include <algorithm>#include <sstream>#define CLR(x) memset(x,0,sizeof(x))#define OFF(x) memset(x,-1,sizeof(x))#define MEM(x,a) memset((x),(a),sizeof(x))#define ALL(x) x.begin(),x.end()#define AT(i,v) for (auto &i:v)#define For_UVa if (kase!=1) cout << endl#define BUG cout << "I am here" << endl#define lookln(x) cout << #x << "=" << x << endl#define look(x) cout << #x << "=" << x#define SI(a) scanf("%d",&a)#define SII(a,b) scanf("%d%d",&a,&b)#define SIII(a,b,c) scanf("%d%d%d",&a,&b,&c)#define Lson l,mid,rt<<1#define Rson mid+1,r,rt<<1|1#define Root 1,n,1#define BigInteger bignconst int MAX_L=2005;// For BigIntegerconst int INF_INT=0x3f3f3f3f;const long long INF_LL=0x7fffffff;const int MOD=1e9+7;const double eps=1e-9;const double pi=acos(-1);typedef long long  ll;using namespace std;const int N=300;int a[N][N];int dp[N][N];int n;int main(){#ifdef LOCAL    freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin);//  freopen("C:\\Users\\john\\Desktop\\out.txt","w",stdout);#endif    int T_T;    int kase=1;    for (int kase=scanf("%d",&T_T);kase<=T_T;kase++){        scanf("%d",&n);        for (int i=1;i<=n;i++) {            for (int j=1;j<=i;j++) {                scanf("%d",&a[i][j]);//              cout << a[i][j] << ' ';            }//          cout << endl;;        }        for (int i=n+1;i<=2*n-1;i++){            for (int j=1;j<=2*n-i;j++) {                scanf("%d",&a[i][j]);//              cout << a[i][j] << ' ';            }//          cout << endl;        }        CLR(dp);        dp[1][1]=a[1][1];        for (int i=2;i<=n;i++) {            for (int j=1;j<=i;j++){                if (j==1) dp[i][j]=dp[i-1][j]+a[i][j];                else if (j==i) dp[i][j]=dp[i-1][j-1]+a[i][j];                else {                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-1])+a[i][j];                }            }        }        for (int i=n+1;i<=2*n-1;i++) {            for (int j=1;j<=2*n-i;j++){                dp[i][j]=max(dp[i-1][j],dp[i-1][j+1])+a[i][j];            }        }        printf("Case %d: %d\n",kase,dp[2*n-1][1]);    }    return 0;}