DFS水题(HDU2616）

Kill the monster

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1635    Accepted Submission(s): 1117

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

 

Sample Output

32-1
import java.awt.RenderingHints.Key;import java.io.BufferedReader;import java.io.IOException;import java.io.InputStreamReader;import java.util.ArrayList;import java.util.Arrays;import java.util.List;import java.util.Queue;import java.util.Scanner;import java.util.Stack;import java.util.TreeSet;public class Main {    static int n, m, sum, max = 10000515;    static boolean d[];    static int shu[][];    public static void main(String[] args) {        Scanner in = new Scanner(System.in);        while (in.hasNext()) {            sum=0;            max=10000515;            n = in.nextInt();            m = in.nextInt();            shu = new int[n][2];            d = new boolean[n];            for (int i = 0; i < n; i++) {                shu[i][0] = in.nextInt();                shu[i][1] = in.nextInt();            }            dfs();            if (max == 10000515) {                System.out.println("-1");            } else {                System.out.println(max);            }        }    }    public static void dfs() {        for (int i = 0; i < n; i++) {            if (d[i] == false) {                d[i] = true;                int s = 0;                if (m <= shu[i][1]) {                    s = shu[i][0] * 2;                } else {                    s = shu[i][0];                }                m -= s;                sum++;                if (m <= 0) {                    if (sum < max) {                        max = sum;                    }                }                                dfs();                m += s;                d[i] = false;                sum--;            }        }    }}