【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=1; i<=l; ++i)
            scanf("%d", &A[i]);
        for (i=1; i<=l; ++i)
            scanf("%d", &B[i]);
        memset(dp, 0x3f, sizeof(dp));
        /*
        for (j=0; j<MAXM; ++j)
            dp[0][j] = INF;
        */
        dp[0][0] = 0;
        int p = 0;
        for (k=1; k<=n; ++k) {
            for (i=1; i<=l; ++i) {
                ++p;
                for (j=0; j<GATE; ++j) {
                    // no speed
                    if (j)
                        dp[p][j] = min(dp[p][j], dp[p-1][j-1]+A[i]);
                    // speed
                    if (j+5<GATE)
                        dp[p][j] = min(dp[p][j], dp[p-1][j+5]+B[i]);
                }
                // check the j=10 because, it can be turn from 14
                dp[p][10] = min(dp[p][10], dp[p-1][14]+A[i]);
            }
        }
        int ans = INF;
        i = l*n;
        for (j=0; j<GATE; ++j)
            ans = min(ans, dp[i][j]);
        printf("%d
", ans);
    }

    return 0;
}