hdu3008_Warcraft

http://acm.hdu.edu.cn/showproblem.php?pid=3008

这道题，从一开始的读错题意到敲代码时犯了很多很低级的错误，纠结了我好久好久，最后还是去看了别人的解题报告，发现自己有个地方想错了。。还从大牛的题解里接触到了try...catch...，果断是个很方便的货~~


死亡时间是一定的，因为每秒都会减少q点LifeValue，所以在死之前把boss给KO掉就行了~~要不然就是My god了。。。

f[ i ][ j ]表示第 i 秒结束时，MagicValue为 j，能消耗boss的最大LifeValue：

      int temp = min(100, j-a[k]+t)    //因为每次回血后MagicValue最多不会超过100

      f[i][temp] = max(f[i][temp],  f[i-1][j] + b[k])   //一开始我把方程错写为：f[i][j] = max(f[i][j],  f[i-1][temp] + b[k])

01 #include <cstdio>
02 #include <cstring>
03 #include <algorithm>
04 using namespace std;
05
06 const int N = 110;
07 int f [N ][N ], a [N ], b [N ];
08
09 int main()
10 {
11     int n , t , q;
12     while( scanf( "%d%d%d" , &n , & t , & q) != EOF && n + t + q)
13     {
14         memset( a , 0 , sizeof( a));
15         memset(b , 0 , sizeof(b));
16         a [ 0 ] = 0;         //把ordinary attack当做第一种攻击
17         b [ 0 ] = 1;
18         for ( int i = 1; i <= n; i ++)
19             scanf( "%d%d" , & a [ i ], &b [ i ]);
20               
21         for ( int i = 0; i < N; i ++)
22             for ( int j = 0; j < N; j ++)
23                 f [ i ][ j ] = - 99999999;      //第一个状态只能由f[0][100]转移过来得到
24         f [ 0 ][ 100 ] = 0;
25        
26         try
27         {
28             for ( int i = 1; ( i - 1) * q < 100; i ++)   
29                 for ( int j = 0; j <= 100; j ++)
30                     for ( int k = 0; k <= n; k ++)
31                         if ( j >= a [ k ])
32                         {
33                             int temp = min( 100 , j - a [ k ] + t);
34                             f [ i ][ temp ] = max( f [ i ][ temp ], f [ i - 1 ][ j ] + b [ k ]);
35                             if ( f [ i ][ temp ] >= 100)
36                                 throw i;
37                         }
38             printf( "My god \n ");
39         }
40         catch ( int ans)
41         {
42             printf( "%d \n " , ans);
43         }
44     }
45 }