洛谷1052（路径压缩后简单dp）

同POJ3744写法都是一样的。

距离太长无意义可以压缩，注意不是随便压的，想一想可以跟%T发生关系。

#include <cstdio>
#include <cctype>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 2500;
int L, S, T, M, pos[101], a[maxn], f[maxn];

int main() {
    scanf("%d %d %d %d", &L, &S, &T, &M);
    for (int i = 1; i <= M; i++)
        scanf("%d", &pos[i]);
    sort(pos + 1, pos + 1 + M);
    for (int i = 1; i <= M; i++) {
        if (pos[i] - pos[i - 1] > T) {
            for (int j = M; j >= i; --j)
                pos[j] -= pos[i] - pos[i - 1] - T - (pos[i] - pos[i - 1]) % T;
        }
    }

    for (int i = 1; i <= M; i++)    a[pos[i]] = 1;
    L = pos[M] + 1;
    for (int i = 1; i <= L; i++)    f[i] = 101;
    for (int i = 0; i < L; i++) {
        for (int j = S; j <= T; j++) {
            int arrive = min(L, i + j);
            f[arrive] = min(f[arrive], f[i] + a[arrive]);
        }
    }

    printf("%d\n", f[L]);
    return 0;
}