The Triangle POJ 1163 递推与记忆化搜索

7
3   8
8   1   0
2   7   4   4
4   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

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

Sample Output

30
递推代码O(n

²

)

//#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cmath>
#include<cstring>
#include <algorithm>
#include <queue>
#include<map>
using namespace std;
typedef long long ll;
const ll inf = 1e13;
const int mod = 1000000007;
const int mx = 150; //check the limits, dummy
typedef pair<int, int> pa;
const double PI = acos(-1);
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
#define swa(a,b) a^=b^=a^=b
#define re(i,a,b) for(int i=(a),_=(b);i<_;i++)
#define rb(i,a,b) for(int i=(b),_=(a);i>=_;i--)
#define clr(a) memset(a, 0, sizeof(a))
#define lowbit(x) ((x)&(x-1))
#define mkp make_pai
//void sc(int& x) { scanf("%d", &x); }void sc(int64_t& x) { scanf("%lld", &x); }void sc(double& x) { scanf("%lf", &x); }void sc(char& x) { scanf(" %c", &x); }void sc(char* x) { scanf("%s", x); }
int n, m, k,ans;
int high[mx];
int a[mx][mx], dp[mx][mx];
int main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    while (~scanf("%d",&n))
    {
        clr(dp),clr(a);
        re(i, 1, n + 1)re(j,1,i+1)scanf("%d",&a[i][j]);
        re(j, 1, n + 1)dp[n][j] = a[n][j];
        for (int i = n - 1; i >= 1; i--)
            for (int j = 1; j <= i; j++)
                dp[i][j] = a[i][j] + max(dp[i + 1][j], dp[i + 1][j + 1]);        
        printf("%d\n",dp[1][1]);
    }
    return 0;
}

下面用递推+记忆化搜索也是O（n²)

//#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cmath>
#include<cstring>
#include <algorithm>
#include <queue>
#include<map>
using namespace std;
typedef long long ll;
const ll inf = 1e13;
const int mod = 1000000007;
const int mx = 150; //check the limits, dummy
typedef pair<int, int> pa;
const double PI = acos(-1);
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
#define swa(a,b) a^=b^=a^=b
#define re(i,a,b) for(int i=(a),_=(b);i<_;i++)
#define rb(i,a,b) for(int i=(b),_=(a);i>=_;i--)
#define clr(a) memset(a, -1, sizeof(a))
#define lowbit(x) ((x)&(x-1))
#define mkp make_pai
//void sc(int& x) { scanf("%d", &x); }void sc(int64_t& x) { scanf("%lld", &x); }void sc(double& x) { scanf("%lf", &x); }void sc(char& x) { scanf(" %c", &x); }void sc(char* x) { scanf("%s", x); }
int n, m, k,ans;
int high[mx];
int a[mx][mx], dp[mx][mx];
int dfs(int i, int j) {
    if (i == n)return a[i][j];
    if (dp[i][j] >= 0)return dp[i][j];
    return dp[i][j] = max(dfs(i + 1, j), dfs(i + 1, j + 1)) + a[i][j];
}
int main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    while (~scanf("%d",&n))
    {
        clr(dp);
        re(i, 1, n + 1)re(j,1,i+1)scanf("%d",&a[i][j]);
        printf("%d\n",dfs(1,1));
    }
    return 0;
}