POJ Cow Exhibition
题目链接:Click Here~
题目意思自己看吧。
算法分析:
对我来想是没有想到,最后看别人的博客才知道的。要把当中的一个条件当作体积。由于两个条件都存在负数,所以还要先保证最后不会再体积中出现负数的情况。这个easy想到就是给其加上一个题目负数的上限就好了。还有的就是当中的正负出现会影响计算时候的正逆顺序。细节自己看吧。我也不太懂得讲。
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std; const int MAXN = 200005;
const int INF = ~0U >> 2;
int dp[MAXN],s[105],f[105];
int main()
{
int n;
while(~scanf("%d",&n))
{
int V = 100000;
for(int i = 0;i < n;++i)
scanf("%d%d",&s[i],&f[i]);
for(int i = 0;i < MAXN;++i)
dp[i] = -INF;
dp[V] = 0;
int maxv,minv;
maxv = minv = V;
for(int i = 0;i < n;++i){
if(s[i] > 0)
for(int j = maxv;j >= minv;--j)
dp[j+s[i]] = max(dp[j+s[i]],dp[j]+f[i]);
else
for(int j = minv;j <= maxv;++j)
dp[j+s[i]] = max(dp[j+s[i]],dp[j]+f[i]);
}
int ans = 0;
for(int i = V;i <= maxv;++i){
if(dp[i] >= 0) ans = max(ans,dp[i]+i-V);
}
printf("%d\n",ans);
}
return 0;
}
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