【練習09】簡單動態規劃 1006 數塔

演算法思路：簡單DP，經典的樹塔模型，反向遞推即可。

代碼如下：

//模板開始#include <string>   #include <vector>   #include <algorithm>   #include <iostream>   #include <sstream>   #include <fstream>   #include <map>   #include <set>   #include <cstdio>   #include <cmath>   #include <cstdlib>   #include <ctime>#include <iomanip>#include <queue>#include <string.h>#define SZ(x) (int(x.size()))using namespace std;int toInt(string s){istringstream sin(s); int t; sin>>t; return t;}template<class T> string toString(T x){ostringstream sout; sout<<x; return sout.str();}typedef long long int64;int64 toInt64(string s){istringstream sin(s); int64 t; sin>>t;return t;}template<class T> T gcd(T a, T b){ if(a<0) return gcd(-a, b);if(b<0) return gcd(a, -b);return (b == 0)? a : gcd(b, a % b);}#define LOCAL//模板結束（通用部分）#define MAXN 105int cost[MAXN][MAXN];int dp[MAXN][MAXN];int N;void init(){memset(dp, 0, sizeof(dp));}int main(){#ifdef LOCAL//freopen("shuju.txt", "r", stdin);#endifint cas;cin>>cas;for(int k = 0; k < cas; k++){cin>>N;for(int i = 1; i <= N; i++){for(int j = 1; j <= i; j++){cin>>cost[i][j];}}init();for(int i = N; i >= 1; i--){for(int j = 1; j <= i; j++){dp[i][j] = max(dp[i + 1][j], dp[i + 1][j + 1]) + cost[i][j];}}cout<<dp[1][1]<<endl;}return 0;}