HDOJ - 3578 DP...巧妙构造DP状态-优快云博客

2010年多校联合的一道题...拿到手瞬间想到了POJ1014..以为是个水题..稍微加几个条件直接01背包就ok了..如果K是<=所有数的和的一个偶数..能背出一个K..又能背出一个K/2...则说明K/2是答案..为了找最大的..就做完背包..从总和大小往前扫..找到了就break..找到最优解..立马敲完代码后各种WA~~~~后来发现了Bug...

<wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr>例如： <div style="text-align:center"><wbr>2</wbr></div> <div style="text-align:center"><wbr><wbr>1 2</wbr></wbr></div> <div><wbr>按我开始的思路..能背出2..又能背出1..所以2是答案..但很明显这个数据是找不到符合要求的..然后我又稍加了些判断..自己出了很多数据...没问题..submit还是WA...然后又发现Bug...然后又各种想如何避免Bug..总之一直纠结于如何用01背包来解决...但没有成功..</wbr></div> <div><wbr>今天看了出题人的solution...短短两句话...惊醒梦中人啊...</wbr></div> <div> <div><span style="color:#ff0000; line-height:21px; word-wrap:normal; word-break:normal"><wbr><wbr>典型dp</wbr></wbr></span></div> <div><span style="color:#ff0000; line-height:21px; word-wrap:normal; word-break:normal"><wbr><wbr>dp[i][j]为拿到第i个橘子扁担的两个篮子重量差为j的最优情况，输出dp[n][0]即可。</wbr></wbr></span></div> <div><span style="color:#ff0000; line-height:21px; word-wrap:normal; word-break:normal"><wbr><wbr>小trick注意有重量为0的橘子，如果拿了重量为0的橘子并且两边重量依旧为0，这种情况不是-1。</wbr></wbr></span></div> </div> <div> <span style="color:#ff0000; line-height:21px; word-wrap:normal; word-break:normal"><wbr></wbr></span><wbr>好了..我这个小菜除了前面牢骚一下就不多说了...觉得这种题还是比较典型的....自己还木有这种思想啊~~！！！！</wbr> </div> <div> <textarea readonly name="code" class="cpp"><iostream> #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int t,T,n,wight[101],TotolWight; int GetMax(int a,int b) { return a>b?a:b; } int GetAbs(int x) { return x<0?-x:x; } int DP() { int i,j,a[2][2001],k=0; memset(a,-1,sizeof(a)); a[0][0]=0; for (i=1;i<=n;i++) { k=1-k; for (j=0;j<=TotolWight;j++) a[k][j]=a[1-k][j]; for (j=0;j<=TotolWight-wight[i];j++) if (a[1-k][j]>=0) a[k][j+wight[i]]=GetMax(a[k][j+wight[i]],a[1-k][j]+wight[i]); for (j=0;j<=TotolWight;j++) if (a[1-k][j]>=0) a[k][GetAbs(j-wight[i])]=GetMax(a[k][GetAbs(j-wight[i])],a[1-k][j]+wight[i]); } if ( !a[k][0] && wight[1] ) return -1; return a[k][0]/2; } int main() { freopen("3578.in","r",stdin); freopen("3578.out","w",stdout); scanf("%d",&T); for (t=1;t<=T;t++) { scanf("%d",&n); TotolWight=0; for (int i=1;i<=n;i++) { scanf("%d",&wight[i]); TotolWight+=wight[i]; } sort(wight+1,wight+1+n); printf("Case %d: %d\n",t,DP()); } return 0; } </textarea><br> </div> </wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr>