【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。

 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 #define MAXN 105
 #define MAXM 20
 #define GATE 15
 #define INF 0x3f3f3f3f
 int A[MAXN];
 int B[MAXN];
 int dp[MAXN*MAXN][MAXM];

 int main() {
     int l ,n;
     int i, j, k;

 #ifndef ONLINE_JUDGE
     freopen("data.in", "r", stdin);
 #endif

     while (scanf("%d %d", &l, &n) != EOF) {
         for (i=; i<=l; ++i)
             scanf("%d", &A[i]);
         for (i=; i<=l; ++i)
             scanf("%d", &B[i]);
         memset(dp, 0x3f, sizeof(dp));
         /*
         for (j=0; j<MAXM; ++j)
             dp[0][j] = INF;
         */
         dp[][] = ;
         int p = ;
         for (k=; k<=n; ++k) {
             for (i=; i<=l; ++i) {
                 ++p;
                 for (j=; j<GATE; ++j) {
                     // no speed
                     if (j)
                         dp[p][j] = min(dp[p][j], dp[p-][j-]+A[i]);
                     // speed
                     if (j+<GATE)
                         dp[p][j] = min(dp[p][j], dp[p-][j+]+B[i]);
                 }
                 // check the j=10 because, it can be turn from 14
                 dp[p][] = min(dp[p][], dp[p-][]+A[i]);
             }
         }
         int ans = INF;
         i = l*n;
         for (j=; j<GATE; ++j)
             ans = min(ans, dp[i][j]);
         printf("%d\n", ans);
     }

     return ;
 }