浙江财经大学
信工学院ACM集训队

HDU 2616 Kill the monster(DFS)

本文由 Ocrosoft 于 2016-07-11 11:11:46 发表

Kill the monster

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1289    Accepted Submission(s): 880

Problem Description
There 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.
 

Input
The 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).
 

Output
For 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
	
3 100 10 20 45 89 5 40 3 100 10 20 45 90 5 40 3 100 10 20 45 84 5 40
 

Sample Output
	
3 2 -1
 

Author
yifenfei
 

Source

题意:主角有n种技能,怪兽有m点血。技能的攻击力是a,当怪兽的血量少于b时,该技能能够造成2a伤害。把怪兽血量消耗完算赢,技能用完血量不为0算输(每个技能限使用一次)。

暴力搜索所有技能使用情况。

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <conio.h>
#include <cmath>
#include <algorithm>
#include <list>
#include <stack>
#include <string>
#include <vector>
#include <cstring>
#include <cctype>
#include <queue>
#include <set>
#include <climits>
#define ms(a) memset(a,0,sizeof(a))
using namespace std;
template <typename T>
T read(T& t)
{
    cin>>t;
    return t;
}
int n,m;//skill,health
int ans;
struct S
{
    int a;//attack
    int hd;//health-double
} s[12];
bool use[12];
void dfs(int h,int k)
{
    if(k>=ans)return;
    if(h<=0)
    {
        ans=min(ans,k);
        return;
    }
    for(int i=0; i<n; i++)
    {
        if(use[i])continue;
        use[i]=1;
        dfs(h-(h<=s[i].hd?2*s[i].a:s[i].a),k+1);
        use[i]=0;
    }
}
int main()
{
    while(cin>>n>>m)
    {
        for(int i=0; i<n; i++)cin>>s[i].a>>s[i].hd;
        ans=12;
        ms(use);
        dfs(m,0);
        printf("%d\n",ans==12?-1:ans);
    }
    return 0;
}

 

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