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("**%dn", 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("%dn", ans); int m = read(); while(m--) { int si = read(), mi = read(); pair now = query(si, mi); printf("%d %dn", now.fi, now.se); } return 0; }
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