hdu 2059 龟兔赛跑（DP）

来源：互联网发布：linux 启动网络服务编辑：程序博客网时间：2024/05/20 05:45

思路：用dp[i][0]表示到第i个加油站不加油所用的最小时间，dp[i][1]表示到第i个加油站加油所用的最小时间。将终点设为第n+1个站，则答案为dp[n+1][0].具体细节看代码。

#include<cstdio>#include<cmath>#include<cstring>#include<iostream>using namespace std;typedef long long LL;const LL INF = 100000000000;const double EPS = 1e-9;int a[110];double dp[110][2];int dcmp(double x, double y){    if(fabs(x - y) < EPS) return 0;    return x > y ? 1 : -1;}int main(){    int l, n, c, t;    int vr, vt1, vt2;    while(scanf("%d", &l) == 1){        scanf("%d%d%d%d%d%d", &n, &c, &t, &vr, &vt1, &vt2);        for(int i = 1; i <= n; ++i)            scanf("%d", &a[i]);        memset(dp, 0, sizeof dp);        a[0] = 0, a[n + 1] = l;        for(int i = 1; i <= n + 1; ++i){            dp[i][0] = dp[i][1] = INF;            for(int j = 0; j < i; ++j){                int d = a[i] - a[j];                    dp[i][0] = min(dp[i][0], dp[j][0] + (double)d / vt2);                    if(d > c)                        dp[i][0] = min(dp[i][0], dp[j][1] + (double)c / vt1 + (double)(d - c) / vt2);                    else                        dp[i][0] = min(dp[i][0], dp[j][1] + (double)d / vt1);                    dp[i][1] = min(dp[i][1], dp[j][0] + (double)d / vt2 + t);                    if(d > c)                        dp[i][1] = min(dp[i][1], dp[j][1] + (double)c / vt1 + (double)(d - c) / vt2 + t);                    else                        dp[i][1] = min(dp[i][1], dp[j][1] + (double)d / vt1 + t);            }        }        double t1 = (double)l / vr, t2;        if(t1 > dp[n + 1][0]) printf("What a pity rabbit!\n");        else printf("Good job,rabbit!\n");    }}

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