CodeForces 358D — Dima and Hares

这题要备忘一下，对于简单的偏序关系对应的价值也可以施行dp。

/*
ID:esxgx1
LANG:C++
PROG:cf358D
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int maxn = 3007;
unsigned joy[maxn][3];
unsigned dp[2][2];

int main(void)
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    #endif

    int N;
    scanf("%d", &N);
    for(int i=0; i<N; ++i) scanf("%u", &joy[i][0]);
    for(int i=0; i<N; ++i) scanf("%u", &joy[i][1]);
    for(int i=0; i<N; ++i) scanf("%u", &joy[i][2]);

    int curr = 0;
    dp[0][0] = joy[0][0], dp[0][1] = joy[0][1];
    for(int i=2; i<=N; ++i) {
        // 所处位置是c, 当前要决定b, 若b<c，即 a < b < c(1), b < a < c(0), b < c < a(0)
        dp[!curr][0] = max(dp[curr][0] + joy[i-1][1], dp[curr][1] + joy[i-1][0]);
        // 若 c < b, 即 c < a < b, a < c < b(2),  c < b < a (1)
        dp[!curr][1] = max(dp[curr][0] + joy[i-1][2], dp[curr][1] + joy[i-1][1]);
        curr = !curr;
    }
    printf("%u\n", dp[curr][0]);
    return 0;
}