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Now ,there are some rectangles. The area of these rectangles is 1* x or 2 * x ,and now you need find a big enough rectangle( 2 * m) so that you can put all rectangles into it(these rectangles can‘t rotate). please calculate the minimum m satisfy the condition.
There are some tests ,the first line give you the test number.
Each test will give you a number n (1<=n<=100)show the rectangles
number .The following n rows , each row will give you tow number a and
b. (a = 1 or 2 , 1<=b<=100).
Each test you will output the minimum number m to fill all these rectangles.
2
3
1 2
2 2
2 3
3
1 2
1 2
1 3
7 4
题意:在 2*m 的矩形里面放若干宽度为1和2的矩形,问 m 最小要多大?
题解:首先,对于宽度为2的矩形,我们直接加上就好了,这是它所需要的容积,对于宽度为 1 的矩形,我们长度先对其排序，这样就保证了选的矩形尽可能的可以并排放下,对sum/2做01背包,得到容量为sum/2的
背包能够放的最大容量m,用sum-dp[sum/2]即为所需最小容量。
#include<stdio.h> #include<iostream> #include<string.h> #include<math.h> #include<algorithm> using namespace std; int x[105],y[105],v[105]; int dp[10005]; int main() { int tcase,n; scanf("%d",&tcase); while(tcase--){ scanf("%d",&n); int ans = 0,cnt=0,sum = 0; for(int i=1;i<=n;i++){ scanf("%d%d",&x[i],&y[i]); if(x[i]==2){ ans+=y[i]; }else{ v[++cnt] = y[i]; sum+=v[cnt]; } } memset(dp,0,sizeof(dp)); sort(v+1,v+cnt+1); for(int i=1;i<=cnt;i++){ for(int j=sum/2;j>=v[i];j--){ dp[j] = max(dp[j],dp[j-v[i]]+v[i]); } } int res = ans+sum-dp[sum/2]; printf("%d\n",res); } }
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原文地址：http://www.cnblogs.com/liyinggang/p/5792147.html
