HDU2616 Kill the monster

                                                         Kill the monster

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 637    Accepted Submission(s): 448

Problem DescriptionThere is a mountain near yifenfei’s hometown. On the mountain lived a big monster. As a hero in hometown, yifenfei wants to kill it.

Now we know yifenfei have n spells, and the monster have m HP, when HP <= 0 meaning monster be killed. Yifenfei’s spells have different effect if used in different time. now tell you each spells’s effects , expressed (A ,M). A show the spell can cost A HP to
monster in the common time. M show that when the monster’s HP <= M, using this spell can get double effect. 

InputThe input contains multiple test cases.
Each test case include, first two integers n, m (2<n<10, 1<m<10^7), express how many spells yifenfei has.
Next n line , each line express one spell. (Ai, Mi).(0<Ai,Mi<=m). 

OutputFor each test case output one integer that how many spells yifenfei should use at least. If yifenfei can not kill the monster output -1. 

Sample Input

3 10010 2045 895  403 10010 2045 905 403 10010 2045 845 40

 

Sample Output

32-1

 

Authoryifenfei 

Source奮鬥的年代
 

Recommendyifenfei

解題思路：深度優先搜尋題目，還要注意技能使用的順序，雖然有”最少“等字眼，但也不適合用廣搜，只要把怪獸殺死即可，技能循序不定，要求使用的最少技能數。
注意，技能使用不能超過給定的技能數（這裡讓我WA了一次，傷心），每個技能不能多次使用。技能使用的順序直接影響殺死怪獸所需技能數，因為，A,B差值的原因。

#include<cstdio>#include<cstring>#include<algorithm>int n,M;int map[11];int min1=12;struct node{    int a;    int m;}spell[11];void dfs(int x){    int i,j;    if(M<=0)    //怪獸死了    {        min1=min1<x?min1:x;    //最小技能數儲存        //printf("min1=%d x=%d\n",min1,x);        return ;    }    else if(x==n)//易錯點：x不能大於n   //這樣就不會使用“虛有”的技能啦        return ;    for(i=0;i<n;i++)     //順序哦，也是從素數環那個題目那裡學來的    {        if(!map[i])        {            map[i]=1;            j=spell[i].a;            if(M<=spell[i].m)                j+=spell[i].a;     //雙倍血量，影響加倍哦            M-=j;            dfs(x+1);            map[i]=0;            M+=j;        }    }}int main(){    int i,j;    while(scanf("%d%d",&n,&M)!=EOF)    {        memset(map,0,sizeof(map));        min1=12;        for(i=0;i<n;i++)            scanf("%d%d",&spell[i].a,&spell[i].m);        dfs(0);        if(min1!=12)            printf("%d\n",min1);        else            printf("-1\n");    }    return 0;}