hdu 1494

地址：http://acm.hdu.edu.cn/showproblem.php?pid=1494

题意：中文……

mark：dp[i][j]代表第i个时间段，有j数量的油，油的数量0~14.

代码：

#include <stdio.h>
#include <string.h>
#define N 0x3f3f3f3f

int dp[2][15];
int a[110][2];

int min(int a, int b) {return a < b ? a : b;}

int main()
{
    int l,n,min1;
    int i,j;
    while(~scanf("%d%d", &l, &n))
    {
        for(i = 0; i < l; i++)
            scanf("%d", &a[i][0]);
        for(i = 0; i < l; i++)
            scanf("%d", &a[i][1]);
        memset(dp, 0x3f, sizeof(dp));
        dp[0][1] = a[0][0];
        for(i = 1; i < l*n; i++)
        {
            for(j = 0; j < 15; j++)
                dp[i%2][j] = N;
            if(dp[!(i%2)][5] != N) dp[i%2][0] = dp[!(i%2)][5]+a[i%l][1];
            for(j = 1; j < 15; j++)
            {
                if(j == 10 && dp[!(i%2)][j-1] == N && dp[!(i%2)][j+4] == N) continue;
                else if(j < 10 && dp[!(i%2)][j-1] == N && dp[!(i%2)][j+5] == N) continue;
                else if(j > 10 && dp[!(i%2)][j-1] == N) continue;
                if(j < 10) dp[i%2][j] = min(dp[!(i%2)][j-1]+a[i%l][0], dp[!(i%2)][j+5]+a[i%l][1]);
                else if(j == 10) dp[i%2][j] = min(dp[!(i%2)][j-1], dp[!(i%2)][14])+a[i%l][0];
                else dp[i%2][j] = dp[!(i%2)][j-1]+a[i%l][0];
            }
        }
        min1 = dp[(l*n-1)%2][0];
        for(i = 1; i < 15; i++)
            min1 = min(min1, dp[(l*n-1)%2][i]);
        printf("%d\n", min1);
    }
    return 0;
}