poj 2184 Cow Exhibition(01背包)
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题目链接:http://poj.org/problem?id=2184
a[i]:第i头牛smartness,b[i]:第i头牛funness.
因为题目中存在负数,第一点先消除负数的状态,看题目可知可定义shift=100000为0点。
dp[i][j]:前i头牛总smartness为j的情况下,总的smartness和funness最大的大小。 dp[i][j]=max(dp[i-1][j],dp[i-1][j-a[i]]+a[i]+b[i]);
#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;const int maxn=200005;const int INF=0x3f3f3f3f;const int N=105;int dp[2][maxn],a[N],b[N];int n;int main(){ //freopen("in.txt","r",stdin); while(~scanf("%d",&n)) { int maxx=0; int sum=n*1000; for(int j=0;j<2;j++) for(int i=100000-sum;i<=100000+sum;i++) dp[j][i]=-INF; dp[0][100000]=0; for(int i=1;i<=n;i++) { scanf("%d%d",&a[i],&b[i]); for(int j=100000+sum;j>=100000-sum;j--) { int num=(i-1)%2; if((dp[num][j-a[i]]!=-INF||dp[num][j]!=-INF)&&j-a[i]<=100000+sum&&j-a[i]>=100000-sum) { dp[i%2][j]=max(dp[num][j-a[i]]+a[i]+b[i],dp[num][j]); if(j>=100000&&j-100000<=dp[i%2][j]) maxx=max(maxx,dp[i%2][j]); } } } cout<<maxx<<endl; } return 0;}
另外一种dp思路:http://blog.csdn.net/liuqiyao_01/article/details/8753686。(若每次都从大到小循环,dp会被自身干扰)。
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