CodeForces 489E Hiking (二分+DP)

题意： 一个人在起点0，有n个休息点，每个点有两个数值，分别表示距离起点的距离xi，以及所获得的愉悦值bi，这个人打算每天走L距离，但实际情况不允许他这么做。定义总体失望值val = sum(sqrt(Ri - L)) / sum(bi); 现在要使得val最小（这个人必须要到达最终的节点）。

析：其实这个题并不难，看好这个式子，只要变形一下，sum(sqrt(Ri - L)) - val * sum(bi) = 0，要求val最小，假设val就是最后答案那肯定满足sum(sqrt(Ri - L)) - val * sum(bi) >= 0，

然后就可以二分val，然后用dp进行判断。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

double dp[maxn];
int a[maxn], b[maxn];
int path[maxn];

bool judge(double mid){
  dp[0] = 0;
  for(int i = 1; i <= n; ++i){
    dp[i] = inf;
    for(int j = 0; j < i; ++j){
      double res = sqrt(fabs(a[i] - a[j] - m)) - mid * b[i] + dp[j];
      if(dp[i] > res){
        dp[i] = res;
        path[i] = j;
      }
    }
  }
  return dp[n] >= 0.0;
}

void print(int x){
  if(x == 0)  return ;
  print(path[x]);
  printf("%d ", x);
}

int main(){
  scanf("%d %d", &n, &m);
  for(int i = 1; i <= n; ++i)  scanf("%d %d", a+i, b+i);
  double l = 0.0, r = 1e10;
  for(int i = 0; i < 100; ++i){
    double mid = (l + r) / 2.0;
    if(judge(mid))  l = mid;
    else r = mid;
  }
  print(n);
  printf("\n");
  return 0;
}