题意

题目链接

Sol

咕了一年的题解。。

并不算是很难，只是代码有点毒瘤

\(f[i][j]\)表示从\(i\)号节点出发走了\(2^j\)轮后总的距离

\(da[i][j]\)同理表示\(a\)的距离,\(db[i][j]\)与\(da\)同理

倍增优化一下

注意最后\(a\)可能还会走一次

#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP make_pair
#define fi first
#define se second 
#define Fin(x) {freopen(#x".in","r",stdin);}
#define pb(x) push_back(x)
#define LL long long 
//#define int long long 
using namespace std;
const int MAXN = 1e5 + 10;
LL INF = 2e9 + 10, B = 20;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, jp[MAXN][21], nxa[MAXN], nxb[MAXN], flag[MAXN], h[MAXN];
LL f[MAXN][21], da[MAXN][21], db[MAXN][21];
struct Node {
    LL Hi, id;
    bool operator < (const Node &rhs) const{
        return Hi == rhs.Hi ? h[id] < h[rhs.id] : Hi < rhs.Hi;
    }
};
int get(int x, int y) {
    return abs(h[x] - h[y]);
}
set<Node> s;
void Pre() {
    s.insert((Node) {-INF * 2, 0}); s.insert((Node) {INF * 2, 0});
    s.insert((Node) {-INF * 2 + 1, 0}); s.insert((Node) {INF * 2 + 1, 0});
    s.insert((Node) {h[N], N});
    memset(f, 0x3f, sizeof(f));
    for(int i = N - 1; i >= 1; i--) {
        set<Node> :: iterator y = s.lower_bound((Node) {h[i], i});
        vector<Node> tmp; tmp.clear();
        tmp.push_back((Node) {get(y -> id, i), y -> id}); y++;
        tmp.push_back((Node) {get(y -> id, i), y -> id}); y--; y--;
        tmp.push_back((Node) {get(y -> id, i), y -> id}); y--;
        tmp.push_back((Node) {get(y -> id, i), y -> id});
        sort(tmp.begin(), tmp.end());
        nxa[i] = tmp[1].id;
        nxb[i] = tmp[0].id;
        s.insert((Node) {h[i], i});
        if(tmp[1].id != 0 && tmp[0].id != 0) f[i][0] = tmp[1].Hi + db[tmp[1].id][0];
        jp[i][0] = nxb[nxa[i]];
        da[i][0] = get(i, nxa[i]);
        db[i][0] = get(i, nxb[i]);
    }
    for(int j = 1; j <= B; j++) 
        for(int i = 1; i <= N; i++) {
            if(jp[i][j - 1]) 
				jp[i][j] = jp[jp[i][j - 1]][j - 1], 
				f[i][j] = f[i][j - 1] + f[jp[i][j - 1]][j - 1],
				da[i][j] = f[i][j] == INF ? 0 : da[i][j - 1] + da[jp[i][j - 1]][j - 1]; 	
		}

}
void print() {
    for(int i = 1; i <= N; i++) printf("**%d\n", f[i][0]);
}
Pair Query(int pos, int val) {
    LL a1 = 0, a2 = 0;
    for(int i = B; ~i; i--) 
        if(f[pos][i] <= val) 
			val -= f[pos][i], a1 += da[pos][i], a2 += f[pos][i] - da[pos][i], pos = jp[pos][i];
    if(da[pos][0] <= val) a1 += da[pos][0];
    return MP(a1, a2);   
}
signed main() {
   // freopen("drive3.in", "r", stdin);
 //   freopen("a.out", "w", stdout);
    N = read();
    for(int i = 1; i <= N; i++) h[i] = read(); h[0] = INF;

    Pre();
   // print();
    int X0 = read(), ans = N;
    double tmp = 1e22; 
    for(int i = 1; i <= N; i++) {
        Pair now = Query(i, X0);
        if(now.se == 0) continue;
        if((double)now.fi / now.se < tmp) tmp = (double)now.fi / now.se, ans = i; 
    }
    printf("%d\n", ans);
    int M = read();
    while(M--) {
        int Si = read(), Mi = read();
        Pair now = Query(Si, Mi);
        printf("%d %d\n", now.fi, now.se);
    }
    return 0;
}