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  • uva1336 Fixing the Great Wall

     用到了kase避免memset超时

    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    #include<cassert>
    using namespace std;
    
    const int maxn = 1000 + 5;
    const double INF = 1e30;
    
    struct Section {
      double x, c, dt;
      bool operator < (const Section& rhs) const {
        return x < rhs.x;
      }
    } s[maxn];
    
    int kase, n;
    int vis[maxn][maxn][2];
    double v, x, d[maxn][maxn][2];
    double psdt[maxn]; // prefix sum of dt
    
    // cost accumulated when walking from x1 and x2.
    // section[i~j] are already finished
    double cost(double x1, double x2, int i, int j) {
      double finished_dt = 0;
      assert(i <= j);//not necessary  
      if(i >= 0 && j >= 0) finished_dt += psdt[j] - psdt[i-1];  // -1 -1�����
      return (psdt[n] - finished_dt) * fabs(x2 - x1) / v;
    }
    
    double dp(int i, int j, int p) {        //p=0��ǰ��i,p=1��ǰ��j
      if(i == 1 && j == n) return 0;
      double& ans = d[i][j][p];
      if(vis[i][j][p] == kase) return ans;
      vis[i][j][p] = kase;
    
      ans=INF;
      double x = (p == 0 ? s[i].x : s[j].x);
      if(i > 1) ans = min(ans, dp(i-1, j, 0) + cost(x, s[i-1].x, i, j));
      if(j < n) ans = min(ans, dp(i, j+1, 1) + cost(x, s[j+1].x, i, j));
      return ans;
    }
    
    int main() {
      memset(vis, 0, sizeof(vis));
      while(scanf("%d%lf%lf", &n, &v, &x) == 3 && n) {
        kase++;
        double sumc = 0;
        for(int i = 1; i <= n; i++) {
          scanf("%lf%lf%lf", &s[i].x, &s[i].c, &s[i].dt);
          sumc += s[i].c;
        }
        sort(s+1, s+n+1);
    
        psdt[0] = 0;
        for(int i = 1; i <= n; i++)
          psdt[i] = psdt[i-1] + s[i].dt;
    
        s[0].x = -INF;
        s[n+1].x = INF;
        double ans = INF;
        for(int i = 1; i <= n+1; i++)
          if(x > s[i-1].x && x < s[i].x) {
            if(i > 1)
                ans = min(ans, dp(i-1, i-1, 0) + cost(x, s[i-1].x, -1, -1)); // move left
            if(i <= n)
                ans = min(ans, dp(i, i, 0) + cost(x, s[i].x, -1, -1)); // move right
            break;
          }
        printf("%.0lf
    ", floor(ans + sumc));
      }
      return 0;
    }
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  • 原文地址:https://www.cnblogs.com/lqerio/p/9800686.html
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