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;

}

巴特西

CodeForces 489E Hiking (二分+DP)

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