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| CD |
You have a long drive by car ahead. You have a tape recorder, but unfortunately your best music is on CDs. You need to have it on tapes so the problem to solve is: you have a tape N minutes long. How to choose tracks from CD to get most out of tape space and have as short unused space as possible.
Assumptions:
Program should find the set of tracks which fills the tape best and print it in the same sequence as the tracks are stored on the CD
Any number of lines. Each one contains value N, (after space) number of tracks and durations of the tracks. For example from first line in sample data: N=5, number of tracks=3, first track lasts for 1 minute, second one 3 minutes, next one 4 minutes
Set of tracks (and durations) which are the correct solutions and string ``sum:" and sum of duration times.
5 3 1 3 4 10 4 9 8 4 2 20 4 10 5 7 4 90 8 10 23 1 2 3 4 5 7 45 8 4 10 44 43 12 9 8 2
1 4 sum:5 8 2 sum:10 10 5 4 sum:19 10 23 1 2 3 4 5 7 sum:55 4 10 12 9 8 2 sum:45
这题主要是输出路径有问题
就是把路径记录下来
其他都好写
现在有两种方法
#include<stdio.h> #include<string.h> #include<algorithm> #include<math.h> #include<stdlib.h> #include<queue> #include<iostream> using namespace std; int dp[111000],vis[50][111000]; int f[50],k=0,w[50]; void prime(int n,int m) { if(n==0||m<0) return; if(vis[n][m]==0) prime(n-1,m); else { prime(n-1,m-w[n]); f[k++]=w[n];; } } int main() { int m,n,i,j; while(scanf("%d",&m)!=EOF) { memset(w,0,sizeof(w)); scanf("%d",&n); for(i=1;i<=n;i++) { scanf("%d",&w[i]); } memset(dp,0,sizeof(dp)); memset(vis,0,sizeof(vis)); for(i=1;i<=n;i++) { for(j=m;j>=w[i];j--) { if(dp[j]<dp[j-w[i]]+w[i]) { dp[j]=dp[j-w[i]]+w[i]; vis[i][j]=1; } } } k=0; memset(f,0,sizeof(f)); prime(n,m); for(i=0;i<k;i++) printf("%d ",f[i]); printf("sum:%d\n",dp[m]); } return 0; }
#include<stdio.h> #include<string.h> #include<algorithm> #include<math.h> #include<stdlib.h> #include<queue> #include<iostream> using namespace std; int dp[111000],vis[50][111000]; int main() { int m,n,w[50],i,j; while(scanf("%d",&m)!=EOF) { scanf("%d",&n); for(i=1;i<=n;i++) { scanf("%d",&w[i]); } memset(dp,0,sizeof(dp)); memset(vis,0,sizeof(vis)); for(i=n;i>0;i--) { for(j=m;j>=w[i];j--) { if(dp[j]<dp[j-w[i]]+w[i]) { dp[j]=dp[j-w[i]]+w[i]; vis[i][j]=1; } } } for(i=1,j=dp[m];i<=n&&j>0;i++) { if(vis[i][j]) { printf("%d ",w[i]); j=j-w[i]; } } printf("sum:%d\n",dp[m]); } return 0; }
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原文地址：http://www.cnblogs.com/linliu/p/5041344.html
