FATE
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 17403 Accepted Submission(s): 8172
Problem Description
最近xhd正在玩一款叫做FATE的游戏,为了得到极品装备,xhd在不停的杀怪做任务。久而久之xhd开始对杀怪产生的厌恶感,但又不得不通过杀怪来升完这最后一级。现在的问题是,xhd升掉最后一级还需n的经验值,xhd还留有m的忍耐度,每杀一个怪xhd会得到相应的经验,并减掉相应的忍耐度。当忍耐度降到0或者0以下时,xhd就不会玩这游戏。xhd还说了他最多只杀s只怪。请问他能升掉这最后一级吗?
Input
输入数据有多组,对于每组数据第一行输入n,m,k,s(0 < n,m,k,s < 100)四个正整数。分别表示还需的经验值,保留的忍耐度,怪的种数和最多的杀怪数。接下来输入k行数据。每行数据输入两个正整数a,b(0 < a,b < 20);分别表示杀掉一只这种怪xhd会得到的经验值和会减掉的忍耐度。(每种怪都有无数个)
Output
输出升完这级还能保留的最大忍耐度,如果无法升完这级输出-1。
Sample Input
10 10 1 101 110 10 1 91 19 10 2 101 12 2
Sample Output
0-11
Author
Xhd
Source
解题思路:
本题是二维的01背包问题 dp忍耐度和最多杀多少只怪 每次更新最大值 并且记录下实际用了多少忍耐度 然后与m相减即可求出结果
import java.util.Scanner;
public class Main{
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()) {
int n = scanner.nextInt();
int m = scanner.nextInt();
int k = scanner.nextInt();
int s = scanner.nextInt();
Monster [] arr = new Monster [k];
for (int i = 0; i < arr.length; i++) {
arr[i] = new Monster(scanner.nextInt(), scanner.nextInt());
}
int [][] dp = new int [s+1][m+1];
int [][] du = new int [s+1][m+1];
for (int i = 0; i < k; i++) {
for (int j = 1; j <= s; j++) {
for (int j2 = m; j2 >= arr[i].rennaidu; j2--) {
int v = dp[j-1][j2-arr[i].rennaidu]+arr[i].jingyan;
if (v > dp[j][j2]) {
dp[j][j2] = v;
du[j][j2] = du[j-1][j2-arr[i].rennaidu]+arr[i].rennaidu;
}
}
}
}
int sum = -1;
for (int i = 0; i <= m; i++) {
if (dp[s][i] >= n) {
sum = m-du[s][i];
break;
}
}
System.out.println(sum);
}
}
}
class Monster{
int jingyan;
int rennaidu;
public Monster(int jingyan, int rennaidu) {
this.jingyan = jingyan;
this.rennaidu = rennaidu;
}
}