题目描述
Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).
Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.
有N件物品和一个容量为V的背包。第i件物品的重量是c[i]，价值是w[i]。求解将哪些物品装入背包可使这些物品的重量总和不超过背包容量，且价值总和最大。
输入输出格式
输入格式：
* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di
第一行：物品个数N和背包大小M
第二行至第N+1行：第i个物品的重量C[i]和价值W[i]
输出格式：
* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints
输出一行最大价值。
输入输出样例
4 6
1 4
2 6
3 12
2 7
23
01背包模板
这道题用二维会爆，比如：
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 using namespace std; 7 8 int m,n,w[3403],c[3403],f[3403][12881]; 9 10 int main() 11 { 12 scanf("%d%d",&n,&m); 13 for (int i=1; i<=n; ++i) 14 scanf("%d%d",&w[i],&c[i]); 15 for (int i=1; i<=n; ++i) 16 for (int j=m; j>0; --j) 17 { 18 if (w[i] <= j) 19 f[i][j] = max(f[i-1][j],f[i-1][j-w[i]]+c[i]); 20 else f[i][j] = f[i-1][j]; 21 } 22 printf("%d",f[n][m]); 23 return 0; 24 }
改一下，
这样就好了
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 5 #define maxn 3403 6 #define maxm 12881 7 8 int m,n,w[maxn],c[maxn],f[maxm]; 9 int main() 10 { 11 scanf("%d%d",&n,&m); 12 for (int i=1;i<=n;++i) 13 scanf("%d%d",&w[i],&c[i]); 14 for (int i=1;i<=n;++i) 15 for (int v=m;v>=w[i];--v) 16 if (f[v-w[i]] + c[i] > f[v]) 17 f[v] = f[v-w[i]]+c[i]; 18 printf("%d",f[m]); 19 return 0; 20 }
看得多了，就自然，理解了，嗯，，，╮(╯▽╰)╭AI~~
如果你不开心，那我就把右边
这个帅傻子分享给你吧，
你看，他这么好看，那么深情的望着你，你还伤心吗？
真的！这照片盯上他五秒钟就想笑了。
一切都会过去的。