动态规划——Warcraft

Have you ever played the Warcraft?It doesn't matter whether you have played it !We will give you such an experience.There are so many Heroes in it,but you could only choose one of them.Each Hero has his own skills.When such a Skill is used ,it costs some MagicValue,but hurts the Boss at the same time.Using the skills needs intellegence,one should hurt the enemy to the most when using certain MagicValue. 

Now we send you to complete such a duty to kill the Boss(So cool~~).To simplify the problem:you can assume the LifeValue of the monster is 100, your LifeValue is 100,but you have also a 100 MagicValue!You can choose to use the ordinary Attack(which doesn't cost MagicValue),or a certain skill(in condition that you own this skill and the MagicValue you have at that time is no less than the skill costs),there is no free lunch so that you should pay certain MagicValue after you use one skill!But we are good enough to offer you a "ResumingCirclet"(with which you can resume the MagicValue each seconds),But you can't own more than 100 MagicValue and resuming MagicValue is always after you attack.The Boss is cruel , be careful!

Input

There are several test cases,intergers n ,t and q (0<n<=100，1<=t<=5，q>0) in the first line which mean you own n kinds of skills ,and the "ResumingCirclet" helps you resume t points of MagicValue per second and q is of course the hurt points of LifeValue the Boss attack you each time(we assume when fighting in a second the attack you show is before the Boss).Then n lines follow,each has 2 intergers ai and bi(0<ai,bi<=100).which means using i skill costs you ai MagicValue and costs the Boss bi LifeValue.The last case is n=t=q=0.

Output

Output an interger min (the minimun time you need to kill the Boss)in one line .But if you die(the LifeValue is no more than 0) ,output "My god"!

Sample Input

4 2 2510 520 1030 2876 704 2 2510 520 1030 2877 700 0 0

Sample Output

4My 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!

题意：

英雄和boss初始都是100点血，英雄有100点蓝。

英雄有n个技能，第i个技能消耗a[i]点蓝，伤害是b[i]。

每回合英雄可以选择技能或普通攻击，普通攻击伤害是1但不耗蓝。

英雄每回合还可以恢复蓝量t，回蓝发生在英雄攻击之后。

boss的伤害每回合是固定的q，求英雄杀死boss的最短时间，若不能杀死，输出My god。

思路：设置dp[i][j]表示在i时间，剩余蓝量为j时对boss的最大伤害

则有dp[i][min(100, j+t-a[k])]=max(dp[i][min(100, j+t-a[k])], dp[i-1][j]+b[k]);

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;int n, t, q;int dp[110][110];int a[110];int b[110];int main(){    while(~scanf("%d%d%d", &n, &t, &q)&&(n+t+q))    {        memset(a, 0, sizeof(a));        memset(b, 0, sizeof(b));        memset(dp, -1, sizeof(dp));        for(int i=1; i<=n; i++)            scanf("%d%d", &a[i], &b[i]);        int time=100/q;        if(100%q)            time++;        a[0]=0;     //普通攻击也算作技能        b[0]=1;        for(int i=0; i<=n; i++)            if(a[i]<=100)                dp[1][100-a[i]]=b[i];       //第一回合        for(int i=2; i<=time; i++)            for(int j=0; j<=100; j++)                for(int k=0; k<=n; k++)                {                    if(dp[i-1][j]==-1)     //若之前没出现过此状态就没法决策                        continue;                    if(a[k]<=min(100, j+t))                        dp[i][min(100, j+t-a[k])]=max(dp[i][min(100, j+t-a[k])], dp[i-1][j]+b[k]);                }        bool flag=false;        for(int i=1; i<=time; i++)              //找出最短时间        {            for(int j=0; j<=100; j++)                if(dp[i][j]>=100)                {                    flag=true;                    printf("%d\n", i);                                break;                }            if(flag)                break;        }        if(!flag)            printf("My god\n");    }    return 0;}