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题意  你去打boss  开始你的蓝和血还有boss的血都是100   每秒你先打boss一下  然后boss打你一下你减少q点血   你有n个技能  第i个技能耗蓝a[i]  对boss的伤害为b[i]   普攻伤害为1   而且你每秒回复t点蓝（恢复后不超过100）  求你最少可以多少次打死boss

你最多能打100/q或者100/q+1次   令d[i][j]表示第i秒所剩蓝量为j时boss剩下的最少血量   m为j还未恢复蓝之前的蓝量  j = min (100, m + t)   那么有  d[i][j] = min (d[i][j], d[i - 1][m + a[k]] - b[k]);

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 105;
int a[N], b[N], d[N][N];
int main()
{
    int n, t, q, ans, ti;
    while (scanf ("%d%d%d", &n, &t, &q), n)
    {
        ++n;
        ti = (100 % q) ? (100 / q + 1) : (100 / q);
        a[1] = 0, b[1] = 1;
        for (int i = 2; i <= n; ++i)
            scanf ("%d%d", &a[i], &b[i]);
            
        memset (d, 0x3f, sizeof (d));
        d[0][100] = 100; ans = 0;

        for (int i = 1; i <= ti; ++i)
            for (int m = 0; m <= 100; ++m)
            {
                int j = min (100, m + t);
                for (int k = 1; k <= n; ++k)
                    if (m + a[k] <= 100)
                        d[i][j] = min (d[i][j], d[i - 1][m + a[k]] - b[k]);
                if (d[i][j] <= 0)
                {
                    ans = i, i = ti;
                    break;
                }
            }

        if (ans == 0) printf ("My god\n");
        else printf ("%d\n", ans);
    }
    return 0;
}

Warcraft

4 2 25
10 5
20 10
30 28
76 70
4 2 25
10 5
20 10
30 28
77 70
0 0 0

4
My god

Hint

Hint:
When fighting,you can only choose one kind of skill or just to use the  ordinary attack in the whole second,the ordinary attack costs the Boss 1 
 points of LifeValue,the Boss can only use ordinary attack which costs a whole second at a time.Good Luck To You! 
 

HDU 3008 Warcraft (DP),布布扣,bubuko.com

HDU 3008 Warcraft (DP)

标签：acm   hdu   dp   动态规划   

原文地址：http://blog.csdn.net/iooden/article/details/38661685