POJ 1163 The Triangle DP

The Triangle

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 41680 Accepted: 25205

Description

73   88   1   02   7   4   44   5   2   6   5(Figure 1)

Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right. 

Input

Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99.

Output

Your program is to write to standard output. The highest sum is written as an integer.

Sample Input

573 88 1 0 2 7 4 44 5 2 6 5

Sample Output

30

Source

IOI 1994

DPDPDP

ACcode：

#pragma warning(disable:4786)//使命名长度不受限制#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈#include <map>#include <set>#include <queue>#include <cmath>#include <stack>#include <cctype>#include <cstdio>#include <cstring>#include <stdlib.h>#include <iostream>#include <algorithm>#define rd(x) scanf("%d",&x)#define rd2(x,y) scanf("%d%d",&x,&y)#define rds(x) scanf("%s",x)#define rdc(x) scanf("%c",&x)#define ll long long int#define maxn 101#define mod 1000000007#define INF 0x3f3f3f3f //int 最大值#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)#define MT(x,i) memset(x,i,sizeof(x))#define PI  acos(-1.0)#define E  exp(1)using namespace std;int a[maxn][maxn];int dp[maxn][maxn],n;int slove(int x,int y){    if(dp[x][y]!=-1)        return dp[x][y];//避免重复计算    if(x==n) return dp[x][y]=a[x][y];//初始化最后一行    return dp[x][y]=a[x][y]+max(slove(x+1,y),slove(x+1,y+1));}int main(){    while(rd(n)!=EOF){        MT(dp,-1);//初始化dp        FOR(i,1,n)            FOR(j,1,i)                rd(a[i][j]);        printf("%d\n",slove(1,1));    }    return 0;}/*573 88 1 0 2 7 4 44 5 2 6 5*/