Input

Input is given from Standard Input in the following format:

$\mathrm{NN}$
${\mathrm{a1a1}}^{}$ ${\mathrm{b1b1}}^{}$ ${\mathrm{c1c1}}^{}$
${\mathrm{a2a2}}^{}$ ${\mathrm{b2b2}}^{}$ ${\mathrm{c2c2}}^{}$
$::$
${\mathrm{aNaN}}^{}$ ${\mathrm{bNbN}}^{}$ ${\mathrm{cNcN}}^{}$

Output

Print the maximum possible total points of happiness that Taro gains.

Sample Input 1 Copy

3
10 40 70
20 50 80
30 60 90

Sample Output 1 Copy

210

If Taro does activities in the order C, B, C, he will gain $\mathrm{70+50+90=21070+50+90=210}$ points of happiness.

Sample Input 2 Copy

1
100 10 1

Sample Output 2 Copy

100

Sample Input 3 Copy

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

Sample Output 3 Copy

46

Taro should do activities in the order C, A, B, A, C, B, A

題意：裸的DP題意，題面單詞很簡單。

思路：下一個狀態選上一個狀態中的不和自己相同的那兩個中的最大值。

轉移方程可以見程式碼。

我的AC程式碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define gg(x) getInt(&x)
using namespace std;
typedef long long ll;
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll n;
ll a[maxn];
ll b[maxn];
ll c[maxn];
ll dp[maxn][5];

int main()
{
    gbtb;
    cin>>n;
    repd(i,1,n)
    {
        cin>>a[i]>>b[i]>>c[i];
    }
    dp[1][1]=a[1];
    dp[1][2]=b[1];
    dp[1][3]=c[1];
    repd(i,2,n)
    {
        dp[i][1]+=max(dp[i-1][2],dp[i-1][3])+a[i];
        dp[i][2]+=max(dp[i-1][1],dp[i-1][3])+b[i];
        dp[i][3]+=max(dp[i-1][1],dp[i-1][2])+c[i];
    }
    cout<<max(dp[n][3],max(dp[n][1],dp[n][2]));
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}