题目链接：uva 10671 - Grid Speed

题目大意：给出N，表示在一个N*N的网格中，每段路长L，如今给出h，v的限制速度，以及起始位置sx，sy，终止位置ex，ey，时间范围st，et，车仅仅走最短路，问说在范围内最快到达和耗油量最小的情况下时间和耗油量。

解题思路：dp[x][y][t]表示在x，y这一点，时间为t的耗油量最小为dp[x][y][t]，vis[x][y][t]表示该情况是否可达。dx，dy表示由起点向终点移动的方向。由于时间为L?60v, 由于v是5的倍数，而且小于50.通过约分能够得知，底数有可能剩余2，3，5，7.为了保证时间为整数，所以我们乘上210.可是由于L中可能包括这些因子，所以仅仅要扩大Lgcd(L,210)

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
const int maxn = 15;
const int maxt = 210005;
const double INF = 0x3f3f3f3f3f3f3f;
const double eps = 1e-9;

int h[maxn], v[maxn], vis[maxn][maxn][maxt];
double dp[maxn][maxn][maxt], mintval, minv;
int n, m, mint, minvtime;

int N, L, sx, sy, ex, ey, st, et, dx, dy;

int gcd (int a, int b) {
    return b == 0 ? a : gcd(b, a%b);
}

void init () {
    scanf("%d%d", &N, &L);
    for (int i = 1; i <= N; i++)
        scanf("%d", &h[i]);
    for (int i = 1; i <= N; i++)
        scanf("%d", &v[i]);
    scanf("%d%d%d%d%d%d", &sx, &sy, &ex, &ey, &st, &et);

    m = 210 / gcd(L, 210);

    st *= m;
    et *= m;
    dx = (sx > ex ? -1 : 1);
    dy = (sy > ey ? -1 : 1);
}

inline double cal (int x) {
    return ((double)L / (80 - 0.03 * x * x));
}

void move (int x, int y, int t) {
    int p = x + dx;
    int q = y + dy;
    //printf("%d %d %d %lf\n", x, y, t, dp[x][y][t]);

    if (p > 0 && p <= N) {
        for (int i = 5; i <= h[y]; i += 5) {
            int ti = t + m * L * 60 / i;
            if (ti > et)
                continue;

            if (vis[p][y][ti])
                dp[p][y][ti] = min(dp[p][y][ti], dp[x][y][t] + cal(i));
            else {
                dp[p][y][ti] = dp[x][y][t] + cal(i);
                vis[p][y][ti] = 1;
            }
        }
    }

    if (q > 0 && q <= N) {
        for (int i = 5; i <= v[x]; i += 5) {
            int ti = t + m * L * 60 / i;
            if (ti > et)
                continue;

            if (vis[x][q][ti])
                dp[x][q][ti] = min(dp[x][q][ti], dp[x][y][t] + cal(i));
            else {
                dp[x][q][ti] = dp[x][y][t] + cal(i);
                vis[x][q][ti] = 1;
            }
        }
    }
}

void solve () {
    memset(vis, 0, sizeof(vis));
    dp[sx][sy][0] = 0;
    vis[sx][sy][0] = 1;

    for (int i = sx; i != ex + dx; i += dx) {
        for (int j = sy; j != ey + dy; j += dy) {
            for (int t = 0; t < et; t++) {
                if (vis[i][j][t])
                    move(i, j, t);
            }
        }
    }

    mint = -1;
    minv = INF;
    for (int t = st; t <= et; t++) {
        if (vis[ex][ey][t]) {
            //printf("%d %lf\n", (int)((double)t / n + 1 - eps), dp[ex][ey][t]);
            if (mint == -1) {
                mint = (int)((double)t / m + 1 - eps);
                mintval = dp[ex][ey][t];
            }

            if (dp[ex][ey][t] < minv) {
                minv = dp[ex][ey][t];
                minvtime = (int)((double)t / m + 1 - eps);
            }
        }
    }
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int i = 1; i <= cas; i++) {
        init();
        solve();
        printf("Scenario %d:\n", i);
        if (mint == -1)
            printf("IMPOSSIBLE\n");
        else {
            printf("The earliest  arrival: %d minutes, fuel %.2lf gallons\n", mint, mintval );
            printf("The economical travel: %d minutes, fuel %.2lf gallons\n", minvtime, minv);
        }
    }
    return 0;
}