首页 > ACM题库 > HDU-杭电 > Hdu 1449 Spiderman-动态规划[解题报告] C++
2013
12-11

Hdu 1449 Spiderman-动态规划[解题报告] C++

Spiderman

问题描述 :

Dr. Octopus kidnapped Spiderman’s girlfriend M.J. and kept her in the West Tower. Now the hero, Spiderman, has to reach the tower as soon as he can to rescue her, using his own weapon, the web.From Spiderman’s apartment, where he starts, to the tower there is a straight road. Alongside of the road stand many tall buildings, which are definitely taller or equal to his apartment. Spiderman can shoot his web to the top of any building between the tower and himself (including the tower), and then swing to the other side of the building. At the moment he finishes the swing, he can shoot his web to another building and make another swing until he gets to the west tower. Figure-1 shows how Spiderman gets to the tower from the top of his apartment – he swings from A to B, from B to C, and from C to the tower. All the buildings (including the tower) are treated as straight lines, and during his swings he can’t hit the ground, which means the length of the web is shorter or equal to the height of the building. Notice that during Spiderman’s swings, he can never go backwards.

You may assume that each swing takes a unit of time. As in Figure-1, Spiderman used 3 swings to reach the tower, and you can easily find out that there is no better way.

输入:

The first line of the input contains the number of test cases K (1 <= K <= 20). Each case starts with a line containing a single integer N (2 <= N <= 5000), the number of buildings (including the apartment and the tower). N lines follow and each line contains two integers Xi, Yi, (0 <= Xi, Yi <= 1000000) the position and height of the building. The first building is always the apartment and the last one is always the tower. The input is sorted by Xi value in ascending order and no two buildings have the same X value.

输出:

For each test case, output one line containing the minimum number of swings (if it’s possible to reach the tower) or -1 if Spiderman can’t reach the tower.

样例输入:

2
6
0 3
3 5
4 3
5 5
7 4
10 4
3
0 3
3 4
10 4

样例输出:

3
-1


题意: 蜘蛛侠去救女友, 每次用web射在楼顶, 然后经过一个扇形弧线, 继续发射web, 知道最右端
的west tower, 现在要求出用最少的跳跃次数到达west tower.

解题思路:
1. 题意很清晰, 每次蜘蛛侠的射出web的长度是小于等于下一个建筑的高度.
即: 距离下个建筑最大距离: dist[i] = sqrt( y[i]*y[i] – (y[i]-y[1])*(y[i]-y[1]) );
从横坐标为j的位置要越过建筑i: x[i]-j <= dist[i];
然后经过一个扇形弧线达到: 2*x[i]-j的位置.
2. 设dp[i]:表示到达横坐标为i的位置需要最少的跳跃次数.
动态方程: dp[ 2*x[i]-j ] = min( dp[ 2*x[i]-j ], dp[j]+1 );

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 2000005
#define SIZE 5005
int n;
int x[SIZE], y[SIZE];
int dp[MAX], dist[SIZE];
inline int min(int a, int b)
{
 return a < b ? a : b;
}
int DP()
{
 dp[x[1]] = 0;
 for(int i = 2; i <= n; ++i)
 {
  for(int j = x[i]-1; j >= x[1]; --j)
  {
   if(dp[j] == -1) continue;
   if( (x[i]-j)*(x[i]-j) > dist[i] ) break;
   if( dp[ 2*x[i]-j ] == -1 || dp[ 2*x[i]-j ] > dp[j]+1)
    dp[ 2*x[i]-j ] = dp[j]+1;
   if( 2*x[i]-j >= x[n] &&  (dp[ x[n] ] == -1 || dp[ x[n] ] > dp[ 2*x[i]-j ]) )
    dp[ x[n] ] = dp[ 2*x[i]-j ];
  }
 }
 return dp[ x[n] ];
}
int main()
{
// freopen("input.txt", "r", stdin);
 int caseNum, i;
 scanf("%d",&caseNum);
 while(caseNum--)
 {
  scanf("%d",&n);
  for(i = 1; i <= n; ++i)
  {
   scanf("%d %d",&x[i], &y[i]);
   dist[i] = y[i]*y[i]-(y[i]-y[1])*(y[i]-y[1]);
  }
  memset(dp, -1, sizeof(dp));
  printf("%d\n",DP());
 }
 return 0;
}

转自:http://blog.sina.com.cn/s/blog_77dc9e080101b035.html


  1. 我没看懂题目
    2
    5 6 -1 5 4 -7
    7 0 6 -1 1 -6 7 -5
    我觉得第一个应该是5 6 -1 5 4 输出是19 5 4
    第二个是7 0 6 -1 1 -6 7输出是14 7 7
    不知道题目例子是怎么得出来的

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