bzoj2516 电梯

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=2516

【题解】

状压dp。

$f_{sta,i}$表示状态为sta，当前在第i层的最小花费时间。状态是个三进制表示，0代表没进过电梯；1代表在电梯里；2表示进过电梯，出来了。

然后考虑当前状态转移出去即可。复杂度$O(Cas * 3^n * nm)$，成功垫底。

据说用2个二进制可以更快（逃

# include <vector>
# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e5 + 10;
const int mod = 1e9+7, inf = 1e9;
const int STATUS_SIZE = 59049 + 5;

int n, m, fir, bin[12];
int f[STATUS_SIZE][22], s[12];
struct pa {
    int t, a, b;
    pa() {}
    pa(int t, int a, int b) : t(t), a(a), b(b) {}
}p[12];

vector<int> ps;

# define abs(x) ((x) > 0 ? (x) : -(x))
# define bit(x, i) (((x) / bin[i]) % 3)
# define dist(i, j) (abs(ps[j-1] - ps[i-1]))

inline int getid(int x) {
    return lower_bound(ps.begin(), ps.end(), x) - ps.begin() + 1;
}

inline void gmin(int &a, int b) {
    if(b < a) a = b;
}


inline void sol() {
    cin >> n >> fir;
    ps.clear(); ps.push_back(fir);
    for (int i=1; i<=n; ++i) {
        scanf("%d%d%d", &p[i].t, &p[i].a, &p[i].b);
        ps.push_back(p[i].a);
        ps.push_back(p[i].b);
    }
    sort(ps.begin(), ps.end());
    ps.erase(unique(ps.begin(), ps.end()), ps.end());
    for (int i=1; i<=n; ++i) {
        p[i].a = getid(p[i].a);
        p[i].b = getid(p[i].b);
    }
    fir = getid(fir); m = ps.size();
    for (int sta=0; sta<bin[n]; ++sta) for (int i=1; i<=m; ++i) f[sta][i] = inf;
    for (int i=1; i<=m; ++i) f[0][i] = dist(i, fir);
    for (int sta=0; sta<bin[n]; ++sta) {
        for (int i=1; i<=m; ++i) {
            for (int j=1; j<=n; ++j) {
                if(bit(sta, j-1) == 1) gmin(f[sta+bin[j-1]][p[j].b], max(f[sta][i] + dist(p[j].b, i), p[j].t));
                else if(bit(sta, j-1) == 0) gmin(f[sta+bin[j-1]][p[j].a], max(f[sta][i] + dist(p[j].a, i), p[j].t));
            }
        }
    }
    int ans = inf;
    for (int i=1; i<=m; ++i) gmin(ans, f[bin[n]-1][i]);
    cout << ans << endl;    
}

int main() {
    bin[0] = 1; for (int i=1; i<=10; ++i) bin[i] = bin[i-1]*3;
    int T; cin >> T;
    while(T--) sol();
    return 0;
}