dp --- CSU 1547: Rectangle
2017-03-05 17:14
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Rectangle
Problem's Link: http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1547
Mean:
给你一些宽为1或2 的木板,让你放在一个宽为二的盒子里面,问你这个盒子最短有多长。
analyse:
简单dp,一开始想错了。
Time complexity: O(n)
Source code:
View Code
Problem's Link: http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1547
Mean:
给你一些宽为1或2 的木板,让你放在一个宽为二的盒子里面,问你这个盒子最短有多长。
analyse:
简单dp,一开始想错了。
Time complexity: O(n)
Source code:
// Memory Time // 1347K 0MS // by : crazyacking // 2015-03-29-22.02 #include<map> #include<queue> #include<stack> #include<cmath> #include<cstdio> #include<vector> #include<string> #include<cstdlib> #include<cstring> #include<climits> #include<iostream> #include<algorithm> #define MAXN 10005 #define LL long long using namespace std; int Cas,n,dp[MAXN]; int main(){ cin>>Cas; while(Cas--) { int cnt = 0 ,sum=0; scanf("%d",&n); int ta, tb; memset(dp,0,sizeof dp); dp[0] = 1; int csum = 0 ; for(int i = 1;i <= n; i ++) { scanf("%d %d",&ta,&tb); if(ta == 2 ) sum += tb; else{ csum += tb ; for(int i = csum;i >= 0 ;i -- ) { if(dp[i] != 0 ) { dp[i+tb] = 1; } } } } for(int i = csum /2 ;i >= 0 ;i--) { if(dp[i] != 0 ) { sum += max(i,csum-i); break; } } printf("%d\n",sum); } return 0; }
View Code
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