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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; }
CodeForces 489E Hiking (二分+DP)
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