Escape from the Hell

Posted 2022-10-29 smallocean

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;