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; }
Accepted | 44 KB | 31 ms | 951 |
2014-10-21 12:41:53
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