POJ 2184 Cow Exhibition (dp 转换01背包)
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题目:点击打开链接
题目大意:
有N个物品,每个物品有属性Si和Fi,-1000 <= Si, Fi <= 1000, 每种物品最多只能选一次,问怎样选使得物品的所有Si和Fi属性之和最大,并且要求Si之和与Fi之和都不能下于0.
思路:
这题想了很久都没思路,于是跟前辈请教了下,恍然大悟。把属性Si当做是物品的费用,Fi当做是价值,然后做01背包即可。
代码:
#include<iostream>#include<queue>#include<stack>#include<cstdio>#include<cstring>#include<cmath>#include<map>#include<set>#include<string>#define MP make_pair#define SQ(x) ((x)*(x))typedef long long int64;const double PI = acos(-1.0);const int INF = 0x3f3f3f3f;using namespace std;const int MOD = 1000000007;const int MAXN = 110;const int ADD = 1000*100;int n, m;int S[MAXN], F[MAXN];int f[1000000];int main(){scanf("%d", &n);int sum = 0;int idx = 0;for(int i=0; i<n; ++i){scanf("%d%d", &S[idx], &F[idx]);if(S[idx]<=0 && F[idx]<=0)continue;if(S[idx]>0)sum += S[idx];++idx;}memset(f, -INF, sizeof(f));f[ADD] = 0;for(int i=0; i<idx; ++i){if(S[i]>0){for(int v=sum+ADD; v>=S[i]; --v){f[v] = max(f[v], f[v-S[i]]+F[i]);}}else{for(int v=0; v-S[i]<=sum+ADD; ++v){f[v] = max(f[v], f[v-S[i]]+F[i]);}}}int ans = 0;for(int i=0; i<=sum; ++i)if(f[i+ADD]>=0)ans = max(i+f[i+ADD], ans);printf("%d\n", ans);return 0;}
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