[USACO15JAN]电影移动Moovie Mooving

挺无语的状态压缩题目,太显然了.

\(f[i]\)表示状态为i的最远到达的距离.

然后暴力枚举每一个可行的点.

然后二分一下即可..

/*header*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#define rep(i , x, p) for(int i = x;i <= p;++ i)
#define sep(i , x, p) for(int i = x;i >= p;-- i)
#define gc getchar()
#define pc putchar
#define ll long long
#define mk make_pair
#define fi first
#define se second
using std::min;
using std::max;
using std::swap;
const int maxN = 20 + 3;

inline int gi() {
  int x = 0,f = 1;char c = gc;
  while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
  while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}return x * f;
}

void print(int x) {
  if(x < 0) pc('-') , x = -x;
  if(x >= 10) print(x / 10);
  pc(x % 10 + '0');
}

int f[1048600] , time[maxN], beg[maxN][1007] , num[1048600];

int search(int p , int x) {
    int l = 1 , r = time[p] , mid , ans = -1;
    while(l <= r) {
        mid = (l + r) >> 1;
        if(beg[p][mid] <= x)
            ans = mid , l = mid + 1;
        else r = mid - 1;
    }
    return ans;
}

int main() {
    int n = gi() , L = gi();
    rep(i , 1, n) {
        time[i] = gi();
        int m = gi();
        rep(j , 1, m) beg[i][j] = gi();
    }
    memset(f , 0x3f, sizeof(f));
    f[0] = 0;
    rep(i , 1, (1 << n) - 1) {
        rep(j , 1, n) {
            if(i & (1 << j - 1)) {
                int p = search(j , f[i ^ (1 << (j - 1))]);
                if(p != -1) f[i] = max(f[i] , beg[j][p] + time[j]);
            }
        }
    }
    int ans = 1e9;
    rep(i , 1, (1 << n) - 1) {
    num[i] = num[i - (i & (-i))] + 1;
    if(f[i] >= m) ans = min(ans , num[i]);
    }
    printf("%d\n", ans == 1e9 ? -1 : ans);
    return 0;
}

原文地址：https://www.cnblogs.com/gzygzy/p/10262593.html