uva 11627 Slalom (二分法)

uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2674

　　一道比较明显的二分题，主要是在二分的判断上耗费了一点时间。

代码如下：

typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<PII> VPII;

#define REP(i, n) for (int i = 0; i < (n); i++)
#define REP_1(i, n) for (int i = 1; i <= (n); i++)

VPII gate;
VI ski;
double w;
int v, n;

bool cmp(PII a, PII b) {
    return a > b;
}

void input() {
    int l, h;
    gate.clear();
    scanf("%lf%d%d", &w, &v, &n);
    REP(i, n) {
        scanf("%d%d", &l, &h);
        gate.PB(MPR(h, l));
    }
    gate.PB(MPR(0, 0));
    sort(ALL(gate), cmp);
//    REP(i, SZ(gate)) cout << gate[i].FI << ends << gate[i].SE << endl;
    ski.clear();
    scanf("%d", &n);
    REP(i, n) {
        scanf("%d", &h);
        ski.PB(h);
    }
    ski.PB(inf);
    ski.PB(-inf);
    sort(ALL(ski));
}

bool test(int x) {
    double L = -finf, R = finf;
    REP(i, SZ(gate) - 1) {
        double l = gate[i].SE, r = gate[i].SE + w, d = ((double) gate[i].FI - gate[i + 1].FI) * v / x;
//        cout << L << ends << R << ends << l << ends << r << ends << d << endl;
        if (r < L || R < l) return false;
        L = max(L, l), R = min(R, r);
        L -= d, R += d;
    }
    return true;
}

int bs() {
    int h = 1, t = 1e7, mk = 0, m;
    while (h <= t) {
        m = (h + t) >> 1;
        if (test(m)) mk = m, h = m + 1;
        else t = m - 1;
    }
    return mk;
}

int main() {
//    freopen("in", "r", stdin);
    int T;
    scanf("%d", &T);
    while (T--) {
        input();
        int ans = ski[upper_bound(ALL(ski), bs()) - ski.begin() - 1];
//        printf("%d %d\n", ans, bs());
        if (ans == -inf || ans == inf) puts("IMPOSSIBLE");
        else printf("%d\n", ans);
    }
    return 0;
}

——written by Lyon