洛谷P1081 开车旅行(倍增)

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;
}