Bone Collector

2019-11-08 02:42:46

字體：大中小

供稿：網(wǎng)友

Bone Collector Time Limit: 1000MS Memory Limit: 65536KB PRoblem Description Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave … The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

Input The first line contain a integer T , the number of cases. Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone. Output One integer per line representing the maximum of the total value (this number will be less than 231).

Example Input

1 5 10 1 2 3 4 5 5 4 3 2 1

Example Output

Hint hdoj2602 http://blog.csdn.net/u014231159/article/details/25605411?utm_source=tuicool&utm_medium=referral Author 翻譯（雖然有翻譯，但還是建議大家自己翻譯試試）：http://fanyi.baidu.com/?aldtype=23#en/zh/

think: 這又是一道英文題，其實(shí)自己不太喜歡做英文題，因?yàn)橛械牡胤阶x不懂的感覺(jué)實(shí)在是不好受，但我們都需要挑戰(zhàn)自己，為將來(lái)打下基礎(chǔ)。回歸題目，這是一道背包問(wèn)題，大意就是給出了幾組測(cè)試?yán)樱缓笥纸o出了骨頭的數(shù)量和骨收集器的體積，然后接著是兩行，第一行包含每個(gè)骨頭的價(jià)值，第二行包含每個(gè)骨頭的體積，要求計(jì)算出骨收集器的最大總價(jià)值。代碼實(shí)現(xiàn)：

#include<stdio.h>int max(int a, int b){ if(a>b) return a; else return b;}int main(){ int x, n; int p[100005], m[100050]; int sum[200000]; int t; scanf("%d", &t); while(t--) { scanf("%d %d", &n, &x); int i, j; memset(sum, 0, sizeof(sum)); for(i=0;i<n;i++) { scanf("%d", &m[i]); } for(i=0;i<n;i++) { scanf("%d", &p[i]); } for(i=0;i<n;i++) { for(j=x;j>=0;j--) { if(j>=p[i]) sum[j] = max(sum[j-p[i]]+m[i], sum[j]); } } printf("%d/n", sum[x]); } return 0;}

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