洛谷P2179 骑行川藏

什么毒瘤...

解：n = 1的，发现就是一个二次函数，解出来一个v的取值范围，选最大的即可。

n = 2的，猜测可以三分。于是先二分给第一段路多少能量，然后用上面的方法求第二段路的最短时间。注意剩余能量不足跑完第二段路的时候，返回INF。

正解是啥拉格朗日乘子法，完全搞不倒...

 /**
  * There is no end though there is a start in space. ---Infinity.
  * It has own power, it ruins, and it goes though there is a start also in the star. ---Finite.
  * Only the person who was wisdom can read the most foolish one from the history.
  * The fish that lives in the sea doesn't know the world in the land.
  * It also ruins and goes if they have wisdom.
  * It is funnier that man exceeds the speed of light than fish start living in the land.
  * It can be said that this is an final ultimatum from the god to the people who can fight.
  *
  * Steins;Gate
  */

 #include <bits/stdc++.h>

 const int N = ;

 double k[N], s[N], vv[N], E;
 int n;

 namespace n1 {
     inline void solve() {
         double v = vv[] + sqrt(E / s[] / k[]);
         printf("%.10f\n", s[] / v);
         return;
     }
 }

 namespace n2 {

     inline double cal(double v) {
         double ans = s[] / v;
         double delta = k[] * s[] * (vv[] - v) * (vv[] - v);
         //printf("E - delta = %.10f \n", E - delta);
         double v2 = vv[] + sqrt((E - delta) / s[] / k[]);
         if(v2 < ) return 1e14;
         //printf("v2 %.10f = %.10f + sqrt(%.10f / %.10f / %.10f) \n", v2, vv[2], E - delta, s[2], k[2]);
         //printf("     =  %.10f + %.10f \n", vv[2], sqrt((E - delta) / s[2] / k[2]));
         //printf("cal %.10f ->  %.10f + %.10f / %.10f \n", v, ans, s[2], v2);
         return ans + s[] / v2;
     }

     inline void solve() {

         double l = , r = vv[] + sqrt(E / s[] / k[]);
         for(int i = ; i <= ; i++) {
             double mid = (l + r) / ;
             //printf("l = %.10f r = %.10f \n", l, r);
             double ml = mid - (r - l) / , mr = mid + (r - l) / ;
             double vl = cal(ml), vr = cal(mr);
             if(vl > vr) {
                 l = ml;
             }
             else {
                 r = mr;
             }
         }
         printf("%.10f\n", cal(r));
         return;
     }
 }

 int main() {
     scanf("%d%lf", &n, &E);
     for(int i = ; i <= n; i++) {
         scanf("%lf%lf%lf", &s[i], &k[i], &vv[i]);
     }

     if(n == ) {
         n1::solve();
         return ;
     }

     if(n == ) {
         n2::solve();
         return ;
     }

     return ;
 }

40分代码

学习了一波模拟退火，突然发现这道题可能比较适合乱搞？于是开始疯狂调参最后成功在LOJ和洛谷上A掉了...

考虑如何随机化得出解。我们随机每条路的能量分配即可。

风速为负的每条路有一个能量下限。在此基础上我们把多出来的能量作为自由能量来进行分配。

初始解就是把自由能量均分...之后我们每次随机出两条路a和b，把a的若干能量给b。这里我给的能量是min(a的能量，T * c) * Rand(0, 1)

这里的c是一个参数。于是我们做到了让调整量随着温度的降低而变小。

然后瞎调一波，从0分优化到了100分......中间有很多脑洞大开改参数的过程...

最好玩的是LOJ的AC代码在洛谷上95分，洛谷的AC代码在LOJ上90分...

 // luogu-judger-enable-o2
 #include <bits/stdc++.h>

 const int N = , INF = 0x3f3f3f3f;

 double k[N], s[N], vv[N], E;
 int n;

 namespace n1 {
     inline void solve() {
         double v = vv[] + sqrt(E / s[] / k[]);
         printf("%.10f\n", s[] / v);
         return;
     }
 }

 namespace n2 {

     inline double cal(double v) {
         double ans = s[] / v;
         double delta = k[] * s[] * (vv[] - v) * (vv[] - v);
         double v2 = vv[] + sqrt((E - delta) / s[] / k[]);
         if(v2 < ) return 1e14;
         return ans + s[] / v2;
     }

     inline void solve() {

         double l = , r = vv[] + sqrt(E / s[] / k[]);
         for(int i = ; i <= ; i++) {
             double mid = (l + r) / ;
             //printf("l = %.10f r = %.10f \n", l, r);
             double ml = mid - (r - l) / , mr = mid + (r - l) / ;
             double vl = cal(ml), vr = cal(mr);
             if(vl > vr) {
                 l = ml;
             }
             else {
                 r = mr;
             }
         }
         printf("%.10f\n", cal(r));
         return;
     }
 }

 namespace Fire {
     const double eps = 1e-;
     double T = , dT = 0.999992;
     double nowE[N], temp[N], lm[N];
     int test[N];

     inline int rd(int l, int r) {
         return rand() % (r - l + ) + l;
     }

     inline double Rand() {
         return 1.0 * rand() / RAND_MAX;
     }

     inline double calv(int i, double e) {
         return vv[i] + sqrt(e / k[i] / s[i]);
     }

     inline double calt(int i, double e) {
         return s[i] / (vv[i] + sqrt(e / k[i] / s[i]));
     }

     inline double init() {
         for(int i = ; i <= n; i++) {
             if(vv[i] < -eps) {
                 lm[i] = k[i] * s[i] * vv[i] * vv[i];
                 E -= lm[i];
             }
         }
         double dt = E / n, ans = ;
         for(int i = ; i <= n; i++) {
             nowE[i] = dt;
             ans += calt(i, lm[i] + nowE[i]);
         }
         return ans;
     }

     inline void solve() {
         double ans, fin = 1e14;
         srand();
         for(int A = ; A <= ; A++) {
             ans = init();
             fin = std::min(ans, fin);
             while(T > eps) {
                 /// Random a new solution
                 int a = rd(, n), b = rd(, n);
                 while(a == b) {
                     a = rd(, n), b = rd(, n);
                 }
                 double deltaE = std::min((long double)nowE[a], (long double)T * 1e8) * Rand();
                 temp[a] = nowE[a] - deltaE;
                 temp[b] = nowE[b] + deltaE;

                 double New = ans - calt(a, lm[a] + nowE[a]) - calt(b, lm[b] + nowE[b])
                                  + calt(a, lm[a] + temp[a]) + calt(b, lm[b] + temp[b]);

                 fin = std::min(fin, New);
                 if(New < ans || Rand() < exp((ans - New) / T)) {
                     ans = New;
                     nowE[a] = temp[a];
                     nowE[b] = temp[b];
                 }
                 T = T * dT;
             }
         }
         printf("%.10f\n", fin);
         return;
     }
 }

 int main() {
     scanf("%d%lf", &n, &E);
     for(int i = ; i <= n; i++) {
         scanf("%lf%lf%lf", &s[i], &k[i], &vv[i]);
     }

     if(n == ) {
         n1::solve();
         return ;
     }

     if(n == ) {
         n2::solve();
         return ;
     }

     Fire::solve();
     return ;
 }

洛谷AC代码

 #include <bits/stdc++.h>

 const int N = , INF = 0x3f3f3f3f;

 double k[N], s[N], vv[N], E;
 int n;

 namespace n1 {
     inline void solve() {
         double v = vv[] + sqrt(E / s[] / k[]);
         printf("%.10f\n", s[] / v);
         return;
     }
 }

 namespace n2 {

     inline double cal(double v) {
         double ans = s[] / v;
         double delta = k[] * s[] * (vv[] - v) * (vv[] - v);
         double v2 = vv[] + sqrt((E - delta) / s[] / k[]);
         if(v2 < ) return 1e14;
         return ans + s[] / v2;
     }

     inline void solve() {

         double l = , r = vv[] + sqrt(E / s[] / k[]);
         for(int i = ; i <= ; i++) {
             double mid = (l + r) / ;
             //printf("l = %.10f r = %.10f \n", l, r);
             double ml = mid - (r - l) / , mr = mid + (r - l) / ;
             double vl = cal(ml), vr = cal(mr);
             if(vl > vr) {
                 l = ml;
             }
             else {
                 r = mr;
             }
         }
         printf("%.10f\n", cal(r));
         return;
     }
 }

 namespace Fire {
     const double eps = 1e-;
     double T = , dT = 0.99999;
     double nowE[N], temp[N], lm[N];
     int test[N];

     inline int rd(int l, int r) {
         return rand() % (r - l + ) + l;
     }

     inline double Rand() {
         return 1.0 * rand() / RAND_MAX;
     }

     inline double calv(int i, double e) {
         return vv[i] + sqrt(e / k[i] / s[i]);
     }

     inline double calt(int i, double e) {
         return s[i] / (vv[i] + sqrt(e / k[i] / s[i]));
     }

     inline double init() {
         for(int i = ; i <= n; i++) {
             if(vv[i] < -eps) {
                 lm[i] = k[i] * s[i] * vv[i] * vv[i];
                 E -= lm[i];
             }
         }
         double dt = E / n, ans = ;
         for(int i = ; i <= n; i++) {
             nowE[i] = dt;
             ans += calt(i, lm[i] + nowE[i]);
         }
         return ans;
     }

     inline void solve() {
         double ans, fin = 1e14;
         srand();
         for(int A = ; A <= ; A++) {
             ans = init();
             fin = std::min(ans, fin);
             while(T > eps) {
                 /// Random a new solution
                 int a = rd(, n), b = rd(, n);
                 while(a == b) {
                     a = rd(, n), b = rd(, n);
                 }
                 double deltaE = std::min((long double)nowE[a], (long double)T * 1e11) * Rand();
                 temp[a] = nowE[a] - deltaE;
                 temp[b] = nowE[b] + deltaE;

                 double New = ans - calt(a, lm[a] + nowE[a]) - calt(b, lm[b] + nowE[b])
                                  + calt(a, lm[a] + temp[a]) + calt(b, lm[b] + temp[b]);

                 fin = std::min(fin, New);
                 if(New < ans || Rand() < exp((ans - New) / T)) {
                     ans = New;
                     nowE[a] = temp[a];
                     nowE[b] = temp[b];
                 }
                 T = T * dT;
             }
         }
         printf("%.8f\n", fin);
         return;
     }
 }

 int main() {
     scanf("%d%lf", &n, &E);
     for(int i = ; i <= n; i++) {
         scanf("%lf%lf%lf", &s[i], &k[i], &vv[i]);
     }

     if(n == ) {
         n1::solve();
         return ;
     }

     if(n == ) {
         n2::solve();
         return ;
     }

     Fire::solve();
     return ;
 }

LOJAC代码

调参心得：△T越接近1，就越慢，同时效果越好。初始温度太大可能会超时...