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  • POJ 3176.Cow Bowling

    Cow Bowling
    Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

    Description

    The cows don't use actual bowling balls when they go bowling. They each take a number (in the range 0..99), though, and line up in a standard bowling-pin-like triangle like this: 

              7 

    3 8

    8 1 0

    2 7 4 4

    4 5 2 6 5
    Then the other cows traverse the triangle starting from its tip and moving "down" to one of the two diagonally adjacent cows until the "bottom" row is reached. The cow's score is the sum of the numbers of the cows visited along the way. The cow with the highest score wins that frame. 

    Given a triangle with N (1 <= N <= 350) rows, determine the highest possible sum achievable.

    Input

    Line 1: A single integer, N 

    Lines 2..N+1: Line i+1 contains i space-separated integers that represent row i of the triangle.

    Output

    Line 1: The largest sum achievable using the traversal rules

    Sample Input

    5
    7
    3 8
    8 1 0
    2 7 4 4
    4 5 2 6 5

    Sample Output

    30

    Hint

    Explanation of the sample: 

              7 
    *
    3 8
    *
    8 1 0
    *
    2 7 4 4
    *
    4 5 2 6 5
    The highest score is achievable by traversing the cows as shown above.

     非常明显的dp+贪心

    对于n行的三角形,其一共有 n*(n+1)/2 个数

    dp[left(i)] = max{ dp[left(i) , dp[i] + a[left(i)] }

    dp[left(i+1)+1] = max{ dp[left(i)+1] , dp[i] + a[left(i)+1] }

    其中,left(i)表示i下一层左边的数

    AC代码:GitHub

     1 /*
     2 By:OhYee
     3 Github:OhYee
     4 HomePage:http://www.oyohyee.com
     5 Email:oyohyee@oyohyee.com
     6 Blog:http://www.cnblogs.com/ohyee/
     7 
     8 かしこいかわいい?
     9 エリーチカ!
    10 要写出来Хорошо的代码哦~
    11 */
    12 
    13 #include <cstdio>
    14 #include <algorithm>
    15 #include <cstring>
    16 #include <cmath>
    17 #include <string>
    18 #include <iostream>
    19 #include <vector>
    20 #include <list>
    21 #include <queue>
    22 #include <stack>
    23 #include <map>
    24 using namespace std;
    25 
    26 //DEBUG MODE
    27 #define debug 0
    28 
    29 //循环
    30 #define REP(n) for(int o=0;o<n;o++)
    31 
    32 #define t(n) (((n) * ((n)+1))/2)
    33 
    34 const int maxn = 1005;
    35 int n;
    36 int a[t(maxn)];
    37 int dp[t(maxn)];
    38 
    39 int left(int n) {
    40     int i;
    41     for(i = 0;t(i) < n;i++);
    42     return n + i;
    43 }
    44 
    45 bool Do() {
    46     if(scanf("%d",&n) == EOF)
    47         return false;
    48     for(int i = 1;i <= t(n);i++)
    49         scanf("%d",&a[i]);
    50 
    51     memset(dp,0,sizeof(dp));
    52     for(int i = 0;i <= t(n - 1);i++) {
    53         dp[left(i)] = max(dp[left(i)],dp[i] + a[left(i)]);
    54         dp[left(i) + 1] = max(dp[left(i) + 1],dp[i] + a[left(i) + 1]);
    55     }
    56 
    57     int Max = -1;
    58     for(int i = t(n - 1) + 1;i <= t(n);i++)
    59         Max = max(Max,dp[i]);
    60 
    61     printf("%d
    ",Max);
    62     return true;
    63 }
    64 
    65 int main() {
    66     while(Do());
    67     return 0;
    68 }
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  • 原文地址:https://www.cnblogs.com/ohyee/p/5451752.html
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