POJ_3624Charm Bracelet(01背包)
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 15410 | Accepted: 7028 |
Description
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.
Input
* 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
Output
* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints
Sample Input
4 6 1 4 2 6 3 12 2 7
Sample Output
23
做的0,1背包问题第一题,详解看《背包九讲》。
因为:3405 * 12885 * 4 / 1024 = 171380 > 65536,栈溢出,所以只能将二维的转化为一维来解决。
二维:
#include <iostream> #include <cstring> #include <string> #include <vector> #include <algorithm> using namespace std; typedef struct Point { int w; int d; }Point; Point point[3405]; int dp[3405][12885]; int n, m; void DP() { for(int i = 1; i <= n; i++) { for(int v = 0; v <= m; v++) { if(v >= point[i].w) { dp[i][v] = max(dp[i-1][v], dp[i-1][v-point[i].w] + point[i].d); } else { dp[i][v] = dp[i-1][v]; } } } } int main() { freopen("in.txt", "r", stdin); int x, y; cin >> n >> m; int count = 1; for(int i = 1; i <= n; i++) { cin >> x >> y; point[count].w = x; point[count++].d = y; } memset(dp, 0, sizeof(dp)); DP(); cout << dp[n][m] << endl; return 0; }
一维:
#include<iostream> #include<cstdio> #include<map> #include<cstring> #include<cmath> #include<vector> #include<algorithm> #include<set> #include<string> #include<queue> #include <stack> #define inf 1<<28 const int MAXN=200010; #define pb(x) push_back(x) #define mem(x,y) memset(x,y,sizeof(x)) #define LL long long using namespace std; typedef struct Point { int w; int d; }Point; int main() { freopen("in.txt", "r", stdin); Point point[3405]; int dp[12885] = {0}; int n, m, x, y; cin >> n >> m; for(int i = 1; i <= n; i++) { cin >> x >> y; point[i].w = x; point[i].d = y; } for(int i = 1; i <= n; i++) { for(int v = m; v >= point[i].w; v--) { dp[v] = max(dp[v], dp[v-point[i].w ]+ point[i].d); } } cout << dp[m] << endl; }