【HDOJ】1494 跑跑卡丁车
DP，将能量映射为0~14，注意当选择这圈加速的时候，这圈就不能再储存能量，同时能量14可能转化为10。
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <iostream> 5 #include <algorithm> 6 using namespace std; 7 8 #define MAXN 105 9 #define MAXM 20 10 #define GATE 15 11 #define INF 0x3f3f3f3f 12 int A[MAXN]; 13 int B[MAXN]; 14 int dp[MAXN*MAXN][MAXM]; 15 16 int main() { 17 int l ,n; 18 int i, j, k; 19 20 #ifndef ONLINE_JUDGE 21 freopen("data.in", "r", stdin); 22 #endif 23 24 while (scanf("%d %d", &l, &n) != EOF) { 25 for (i=1; i<=l; ++i) 26 scanf("%d", &A[i]); 27 for (i=1; i<=l; ++i) 28 scanf("%d", &B[i]); 29 memset(dp, 0x3f, sizeof(dp)); 30 /* 31 for (j=0; j<MAXM; ++j) 32 dp[0][j] = INF; 33 */ 34 dp[0][0] = 0; 35 int p = 0; 36 for (k=1; k<=n; ++k) { 37 for (i=1; i<=l; ++i) { 38 ++p; 39 for (j=0; j<GATE; ++j) { 40 // no speed 41 if (j) 42 dp[p][j] = min(dp[p][j], dp[p-1][j-1]+A[i]); 43 // speed 44 if (j+5<GATE) 45 dp[p][j] = min(dp[p][j], dp[p-1][j+5]+B[i]); 46 } 47 // check the j=10 because, it can be turn from 14 48 dp[p][10] = min(dp[p][10], dp[p-1][14]+A[i]); 49 } 50 } 51 int ans = INF; 52 i = l*n; 53 for (j=0; j<GATE; ++j) 54 ans = min(ans, dp[i][j]); 55 printf("%d\n", ans); 56 } 57 58 return 0; 59 }
