Escape from the Hell

Escape from the Hell

[JAG Asia 2016]

容易证明优先选择差值大的更优

对于最后一瓶我们可以枚举

枚举最后一瓶，然后在树状数组上消去它的影响，然后线段树check是否出现被追上的情况，即查询区间最小值。

需要用到两个线段树，因为当二分找到的位置在最后一瓶后面，需要在线段树上消去最后一瓶的影响。

特别注意当差值为负数的时候前缀和就没有单调性了，所以二分要在单调递增区间二分。

#include <bits/stdc++.h>

#define ll long long
using namespace std;
const int maxn = 1e5 + 7;
const ll inf = 0x3f3f3f3f3f3f3f3f;
int n;
int C[maxn];
struct node {
    int a, b;
} s[maxn];

bool cmp(node x, node y) {
    return x.a - x.b > y.a - y.b;
}

ll c[maxn];
ll sum[maxn];
struct tree {
    int l, r;
    ll min1, min2;
} t[maxn << 2];

int lowbit(int x) {
    return x & (-x);
}

ll getsum(int i) {
    ll res = 0;
    while (i > 0) {
        res += c[i];
        i -= lowbit(i);
    }
    return res;
}

void update(int i, ll val) {
    while (i < maxn) {
        c[i] += val;
        i += lowbit(i);
    }
}

void build(int p, int l, int r) {
    t[p].l = l, t[p].r = r;
    if (l == r) {
        t[p].min1 = getsum(l) - sum[l];
        t[p].min2 = getsum(l) - sum[l - 1];
        return;
    }
    int mid = (l + r) >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    t[p].min1 = min(t[p << 1].min1, t[p << 1 | 1].min1);
    t[p].min2 = min(t[p << 1].min2, t[p << 1 | 1].min2);
}

ll ask1(int p, int l, int r) {
    if (l <= t[p].l && r >= t[p].r) return t[p].min1;
    int mid = (t[p].l + t[p].r) >> 1;
    ll val = inf;
    if (l <= mid) val = min(val, ask1(p << 1, l, r));
    if (r > mid) val = min(val, ask1(p << 1 | 1, l, r));
    return val;
}

ll ask2(int p, int l, int r) {
    if (l <= t[p].l && r >= t[p].r) return t[p].min2;
    int mid = (t[p].l + t[p].r) >> 1;
    ll val = inf;
    if (l <= mid) val = min(val, ask2(p << 1, l, r));
    if (r > mid) val = min(val, ask2(p << 1 | 1, l, r));
    return val;
}

int erfen(int z, int y, ll x) {
    int l = z, r = y;
    while (l < r) {
        int mid = (l + r) >> 1;
        if (getsum(mid) >= x)r = mid;
        else l = mid + 1;
    }
    return l;
}

void dubug() {
    for (int i = 1; i <= n; ++i) {
        cout << ask1(1, i, i) << "    " << ask2(1, i, i) << endl;
    }
}

int main() {
    ll L;
    scanf("%d%lld", &n, &L);
    for (int i = 1; i <= n; ++i) {
        scanf("%d%d", &s[i].a, &s[i].b);
    }
    sort(s + 1, s + 1 + n, cmp);
    int k=-1;
    for (int i = 1; i <= n; ++i) {
        if(s[i].a-s[i].b>=0) update(i, s[i].a - s[i].b);
        else if(k==-1){
            k=i-1;
        }
        scanf("%d", &C[i]);
        sum[i] = sum[i - 1] + C[i];
    }
    if(k==-1) k=n;
    build(1, 1, n);
    int minn = n + 1;
    for (int i = 1; i <= n; ++i) {
        if(s[i].a-s[i].b>0) update(i, s[i].b - s[i].a);
        int pp = erfen(1, k + 1, L - s[i].a);
        if (pp != k + 1) {
            if (pp < i) {
                if (ask1(1, 1, pp) > 0) {
                    minn = min(minn, pp + 1);
                }
            } else {
                if (i==1||ask1(1, 1, i - 1) > 0) {
                    if (pp==1||ask2(1, i + 1, pp) - (s[i].a - s[i].b) > 0) {
                        minn = min(minn, pp);
                    }
                }
            }
        }
        if(s[i].a-s[i].b>0)
        update(i, s[i].a - s[i].b);
    }
    if (minn == n + 1) {
        printf("-1\n");
    } else {
        printf("%d\n", minn);
    }
    return 0;
}