POJ2184Cow Exhibition
来源:互联网 发布:青年文明号 网络歌手 编辑:程序博客网 时间:2024/06/04 20:11
一开始准备用二维的背包,且不说空间不够用,二维还开错了....然后瞄了眼题解,发现一维数组可以搞定...下标记录一组费用,数组内容记录一组费用...200000的数组.因为数据有正有负,一维不好解决,于是开了200000*100,超内存.接着用滚动数组解决了.其实可以分正负,一维搞定.贴上自己代码,和一维代码
我的:
#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#define max(a,b) (a)>(b)?(a):(b)#define min(a,b) (a)<(b)?(a):(b)#define PINF 0x7FFFFFFF#define NINF -0x7FFFFFFFint r[200001][2],s[105],f[105],n;int main(){ int i,j,max,k; scanf("%d",&n); for(i=0; i<=200000; i++) for(j=0; j<=1; j++) r[i][j]=NINF; r[100000][0]=0;k=0; for(i=1; i<=n; i++) { scanf("%d%d",&s[i],&f[i]); } for(i=1; i<=n; i++) { for(j=200000; j>=0; j--) { if(j-s[i]<=200000&&j-s[i]>=0&&r[j-s[i]][k]!=NINF&&r[j-s[i]][k]+f[i]>r[j][k]) r[j][1-k]=r[j-s[i]][k]+f[i]; else r[j][1-k]=r[j][k]; } if(i==1)r[100000][0]=NINF; k=1-k; } max=0; for(i=100000; i<=200000; i++) for(j=0; j<=1; j++) { if(r[i][j]>=0&&i-100000+r[i][j]>max)max=i-100000+r[i][j]; } printf("%d\n",max); return 0;}一维:
#include <iostream>using namespace std;const int N=200001, MID = 100000;int n, a, b, l, r, dp[N];//dp的角标代表sumS,值代表sumFint main(){dp[MID] = MID;//a=0, b=0scanf("%d", &n);for(int i = 1; i <= n; i++){scanf("%d %d", &a, &b);l = MID-1000*(i-1)+a;//优化迭代区间r = MID+1000*(i-1)+a;if(a > 0){for(int j = r; j >= l; j--)//a > 0 就从反向迭代if(dp[j-a] || dp[j])dp[j] = max(dp[j], dp[j-a]+b);}else{for(int j = l; j <= r; j++)if(dp[j-a] || dp[j])dp[j] = max(dp[j], dp[j-a]+b);}}int ans = 0;for(int i = MID; i < N; i++)//dp[i]在这个区间存在,sumS >= 0if(dp[i] && dp[i] >= MID)//&& sumF >= 0ans = max(ans, i-MID+dp[i]-MID);printf("%d\n", ans);return 0;}
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