Poj 1925 Spiderman (DP)

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Spiderman sticks the Spiderman to a back-end building from the leftmost building. Then, the Spiderman hangs over and sends the spider silk until it reaches the last building. Ask at least a few splits.

Idea: Skip a building from x-axis J. I must meet the following requirements: (P [I]. x-j) * (p [I]. x-j) <= P [I]. y * P [I]. y-(P [I]. y-P [0]. y) * (p [I]. y-P [0]. Y ). from X coordinate J to X coordinate 2 * P [I] After building I. x-J.

So the state transition equation is: DP [2 * P [I]. x-J] = min (DP [2 * P [I]. x-J], d [J] + 1 ). Because this person follows an arc, the end point of each swing must be the height of the first building.

First, traverse the building, and then traverse the location. The second layer loops look for where he wants to swing, because it is symmetric.

 1 //1925 2 #include <cstdio> 3 #include <string> 4 #include <cstring> 5 #include <iostream> 6  7 using namespace std ; 8  9 struct node10 {11     long long x ;12     long long  y ;13 }p[5120] ;14 int dp[2051532] ;15 16 int main()17 {18     int K ,N;19     scanf("%d",&K) ;20     while(K--)21     {22         scanf("%d",&N) ;23         for(int i = 0 ; i < N ; i++)24             scanf("%I64d %I64d",&p[i].x,&p[i].y) ;25         memset(dp,100,sizeof(dp)) ;26         dp[p[0].x] = 0 ;27         for(int i = 1 ; i < N ; i++)28             for(int j = p[i].x-1 ; j >= p[0].x ; j--)29         {30             long long  dis = (p[i].x-j)*(p[i].x-j)+(p[i].y-p[0].y)*(p[i].y-p[0].y) ;31             if(dis > (long long)(p[i].y*p[i].y))32                 break ;33             dp[2*p[i].x-j] = min(dp[2*p[i].x-j],dp[j]+1) ;34         }35         int ans = 99999999 ;36         for(int i = p[N-1].x ; i <= 2*p[N-1].x ; i++)37             ans = min(ans,dp[i]) ;38         if(ans >= 99999999)39             ans = -1 ;40         printf("%d\n",ans) ;41     }42     return 0 ;43 }

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