lightoj 1004 - Monkey Banana Problem 【dp】

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本文解析了一道经典的算法题目——猴子吃香蕉问题。题目要求在给定的二维钻石形状数组中，找到一条路径使得猴子能吃到最多的香蕉。文章提供了详细的解题思路及AC代码实现。

题目链接：lightoj 1004 - Monkey Banana Problem

1004 - Monkey Banana Problem
PDF (English) Statistics Forum
Time Limit: 2 second(s) Memory Limit: 32 MB
You are in the world of mathematics to solve the great “Monkey Banana Problem”. It states that, a monkey enters into a diamond shaped two dimensional array and can jump in any of the adjacent cells down from its current position (see figure). While moving from one cell to another, the monkey eats all the bananas kept in that cell. The monkey enters into the array from the upper part and goes out through the lower part. Find the maximum number of bananas the monkey can eat.

Input
Input starts with an integer T (≤ 50), denoting the number of test cases.

Every case starts with an integer N (1 ≤ N ≤ 100). It denotes that, there will be 2*N - 1 rows. The ith (1 ≤ i ≤ N) line of next N lines contains exactly i numbers. Then there will be N - 1 lines. The jth (1 ≤ j < N) line contains N - j integers. Each number is greater than zero and less than 215.

Output
For each case, print the case number and maximum number of bananas eaten by the monkey.

Sample Input
Output for Sample Input
2
4
7
6 4
2 5 10
9 8 12 2
2 12 7
8 2
10
2
1
2 3
1
Case 1: 63
Case 2: 5
Note
Dataset is huge, use faster I/O methods.

题意：给定2*N-1行，要求按照数塔的规定走一条路线使得路线上数之和最大。

思路：我们枚举中间的起点，这样就是双树塔了。注意不合法状态不能转移。

AC代码：

//#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <stack>
#define PI acos(-1.0)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define fi first
#define se second
#define ll o<<1
#define rr o<<1|1
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MAXN = 1e6 + 10;
const int pN = 1e6;// <= 10^7
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
void add(LL &x, LL y) { x += y; x %= MOD; }
LL a[110][110], b[110][110];
LL dp1[110][110], dp2[110][110];
int n;
LL Solve(int s) {
    CLR(dp1, 0); dp1[n][s] = a[n][s];
    for(int i = n-1; i >= 1; i-- ){
        for(int j = 1; j <= i; j++) {
            if(dp1[i+1][j+1] && dp1[i+1][j]) {
                dp1[i][j] = max(dp1[i+1][j+1], dp1[i+1][j]) + a[i][j];
            }
            else if(dp1[i+1][j+1]) {
                dp1[i][j] = dp1[i+1][j+1] + a[i][j];
            }
            else if(dp1[i+1][j]) {
                dp1[i][j] = dp1[i+1][j] + a[i][j];
            }
        }
    }
    CLR(dp2, 0); dp2[n][s] = b[n][s];
    for(int i = n-1; i >= 1; i--) {
        for(int j = 1; j <= i; j++) {
            if(dp2[i+1][j+1] && dp2[i+1][j]) {
                dp2[i][j] = max(dp2[i+1][j+1], dp2[i+1][j]) + b[i][j];
            }
            else if(dp2[i+1][j+1]) {
                dp2[i][j] = dp2[i+1][j+1] + b[i][j];
            }
            else if(dp2[i+1][j]) {
                dp2[i][j] = dp2[i+1][j] + b[i][j];
            }
        }
    }
    return dp1[1][1] + dp2[1][1] - a[n][s];
}
int main()
{
    int t, kcase = 1;
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= i; j++) {
                scanf("%lld", &a[i][j]);
            }
        }
        for(int i = n; i >= 1; i--) {
            for(int j = 1; j <= i; j++) {
                if(i == n) {
                    b[i][j] = a[i][j];
                }
                else {
                    scanf("%lld", &b[i][j]);
                }
            }
        }
        LL ans = 0;
        for(int i = 1; i <= n; i++) {
            ans = max(ans, Solve(i));
        }
        printf("Case %d: %lld\n", kcase++, ans);
    }
    return 0;
}