Ural_1031.Railway Ticket (DP)

　　/*思路：定义dp[i]表示到达第i站的最小花费。所以 dp[i] = min(dp[i-1] + Cx, dp[i-1] + Cx, dp[i-3] + Cx...(直到i 和 (i-x) 两站的距离大于L3)；*/

/*My Code：*/

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int N = 10007;
const int inf = 0x7fffffff;

int dis[N];
int dp[N];

int main() {
    //freopen("data.in", "r", stdin);

    int n, S, E, min;
    int l1, l2, l3;
    int c1, c2, c3;
    int i, j, tmp;

    memset(dp, 0, sizeof(dp));
    memset(dis, 0, sizeof(dis));

    scanf("%d%d%d%d%d%d", &l1, &l2, &l3, &c1, &c2, &c3);
    scanf("%d", &n);
    scanf("%d%d", &S, &E);

    if(S > E)    {tmp = S; S = E; E = tmp;}
    for(i = 2; i <= n; i++)
        scanf("%d", &dis[i]);

    for(i = S+1; i <= E; i++) {
        for(min = inf, j = i-1; j >= S; j--) {
            tmp = dis[i] - dis[j];
            if(tmp <= l1)    tmp = dp[j] + c1;
            else if(tmp <= l2 && tmp > l1)    tmp = dp[j] + c2;
            else if(tmp > l2 && tmp <= l3)     tmp = dp[j] + c3;
            else break;
            if(min > tmp)    min = tmp;
        }
        dp[i] = min;
    }
    printf("%d\n", dp[E]);
    return 0;
}