ACdream 1092 EOF女神的打地鼠游戏 —— dp

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题目链接：http://acdream.info/problem?pid=1092

题解：

基础dp，首先要根据每个地鼠出现的时间对他们排序，然后就可以递推，递归（记忆化搜索）。

递推：

StatusAcceptedTime308msMemory1716kBLength1086LangC++Submitted2017-03-24 19:00:04SharedRemoteRunId99339

#include<iostream>//ACdream - 1092 dp 递推#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>#define MAX(a,b) (a>b?a:b)using namespace std;typedef long long LL;int n,v;double dp[1005], ans;struct node{    int x,y,t;    double r;}m[1005];bool cmp(node a, node b){    return a.t<b.t;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&v);        for(int i = 0; i<n; i++)            scanf("%d%d%d%lf",&m[i].x, &m[i].y, &m[i].t,&m[i].r );        sort(m,m+n,cmp);        ans = 0;        memset(dp,0,sizeof(dp));        for(int i = 0; i<n; i++)        {            dp[i] = m[i].r;            for(int j = 0; j<i; j++)            {                double dis1 = sqrt( (m[i].x-m[j].x)*(m[i].x-m[j].x) + (m[i].y-m[j].y)*(m[i].y-m[j].y) );                double dis2 = 1.0*v*(m[i].t-m[j].t);                if(dis2>=dis1)                    dp[i] = MAX(dp[i],m[i].r+dp[j]);            }            ans = MAX(dp[i],ans);        }        printf("%.6lf\n",ans);    }    return 0;}

递归（正向，跟递推一样）StatusAcceptedTime380msMemory1716kBLength1109LangC++Submitted2017-03-24 18:56:00SharedRemoteRunId99338

#include<iostream>//ACdream - 1092#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>#define MAX(a,b) (a>b?a:b)using namespace std;typedef long long LL;int n,v;double dp[1005], ans;struct node{    int x,y,t;    double r;}m[1005];bool cmp(node a, node b){    return a.t<b.t;}double dfs(int k){    if(dp[k]) return dp[k];    dp[k] = m[k].r;    for(int i = 0; i<k; i++)    {        double dis1 = sqrt( (m[k].x-m[i].x)*(m[k].x-m[i].x) + (m[k].y-m[i].y)*(m[k].y-m[i].y) );        double dis2 = 1.0*v*(m[k].t-m[i].t);        if(dis2>=dis1)            dp[k] = MAX(dp[k],dfs(i)+m[k].r);    }    return dp[k];}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&v);        for(int i = 0; i<n; i++)            scanf("%d%d%d%lf",&m[i].x, &m[i].y, &m[i].t,&m[i].r );        sort(m,m+n,cmp);        ans = 0;        memset(dp,0,sizeof(dp));        for(int i = 0; i<n; i++)        {            ans = MAX(ans,dfs(i));        }        printf("%.6lf\n",ans);    }}

递归（反向）：

StatusAcceptedTime328msMemory1716kBLength1236LangC++Submitted2017-03-24 18:54:43SharedRemoteRunId99337

#include<iostream>//ACdream - 1092 dp 递归 #include<cstdio>  #include<cstring>  #include<cstdlib>  #include<cmath>  #include<algorithm>  #define MAX(a,b) (a>b?a:b)    using namespace std;  typedef long long LL;    int n,v;  double dp[1005], ans;    struct node  {      int x,y,t;      double r;  }m[1005];      bool cmp(node a, node b)  {      return a.t<b.t;  }    double dfs(int k)  {      if(dp[k]) return dp[k];        dp[k] = m[k].r;      for(int i = k+1; i<n; i++)      {          double dis1 = sqrt( (m[k].x-m[i].x)*(m[k].x-m[i].x) + (m[k].y-m[i].y)*(m[k].y-m[i].y) );          double dis2 = 1.0*v*(m[i].t-m[k].t);            if(dis2>=dis1)              dp[k] = MAX(dp[k],dfs(i)+m[k].r);      }      return dp[k];  }    int main()  {      int T;      scanf("%d",&T);      while(T--)      {          scanf("%d%d",&n,&v);          for(int i = 0; i<n; i++)              scanf("%d%d%d%lf",&m[i].x, &m[i].y, &m[i].t,&m[i].r );          sort(m,m+n,cmp);            ans = 0;          memset(dp,0,sizeof(dp));          for(int i = 0; i<n; i++)          {              if(!dp[i])              ans = MAX(ans,dfs(i));          }          printf("%.6lf\n",ans);      }  }

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