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    1004 - Monkey Banana Problem
    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. Thejth (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

    数塔变形题目 easy

    /* ***********************************************
    Author        :guanjun
    Created Time  :2016/6/8 10:57:05
    File Name     :1004.cpp
    ************************************************ */
    #include <iostream>
    #include <cstring>
    #include <cstdlib>
    #include <stdio.h>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <set>
    #include <map>
    #include <string>
    #include <math.h>
    #include <stdlib.h>
    #include <iomanip>
    #include <list>
    #include <deque>
    #include <stack>
    #define ull unsigned long long
    #define ll long long
    #define mod 90001
    #define INF 0x3f3f3f3f
    #define maxn 10010
    #define cle(a) memset(a,0,sizeof(a))
    const ull inf = 1LL << 61;
    const double eps=1e-5;
    using namespace std;
    priority_queue<int,vector<int>,greater<int> >pq;
    struct Node{
        int x,y;
    };
    struct cmp{
        bool operator()(Node a,Node b){
            if(a.x==b.x) return a.y> b.y;
            return a.x>b.x;
        }
    };
    
    bool cmp(int a,int b){
        return a>b;
    }
    int a[210][210];
    int dp[210][210];
    int main()
    {
        #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        #endif
        //freopen("out.txt","w",stdout);
        int T,n;
        cin>>T;
        for(int t=1;t<=T;t++){
            cin>>n;
            cle(a);cle(dp);
            for(int i=1;i<=n;i++){
                for(int j=1;j<=i;j++)
                    cin>>a[i][j];
            }
            for(int i=n+1;i<=2*n-1;i++){
                for(int j=1;j<=2*n-i;j++)
                    cin>>a[i][j];
            }
            for(int i=1;i<=n;i++){
                for(int j=1;j<=i;j++){
                    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+1],dp[i-1][j])+a[i][j];
                }
            }
            printf("Case %d: %d
    ",t,dp[2*n-1][1]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/pk28/p/5569464.html
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