poj 2184 Cow Exhibition 01背包变形
来源:互联网 发布:java流程图不包括 编辑:程序博客网 时间:2024/05/21 14:53
点击打开链接链接
Cow Exhibition
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9288 Accepted: 3551
Description
"Fat and docile, big and dumb, they look so stupid, they aren't much
fun..."
- Cows with Guns by Dana Lyons
The cows want to prove to the public that they are both smart and fun. In order to do this, Bessie has organized an exhibition that will be put on by the cows. She has given each of the N (1 <= N <= 100) cows a thorough interview and determined two values for each cow: the smartness Si (-1000 <= Si <= 1000) of the cow and the funness Fi (-1000 <= Fi <= 1000) of the cow.
Bessie must choose which cows she wants to bring to her exhibition. She believes that the total smartness TS of the group is the sum of the Si's and, likewise, the total funness TF of the group is the sum of the Fi's. Bessie wants to maximize the sum of TS and TF, but she also wants both of these values to be non-negative (since she must also show that the cows are well-rounded; a negative TS or TF would ruin this). Help Bessie maximize the sum of TS and TF without letting either of these values become negative.
fun..."
- Cows with Guns by Dana Lyons
The cows want to prove to the public that they are both smart and fun. In order to do this, Bessie has organized an exhibition that will be put on by the cows. She has given each of the N (1 <= N <= 100) cows a thorough interview and determined two values for each cow: the smartness Si (-1000 <= Si <= 1000) of the cow and the funness Fi (-1000 <= Fi <= 1000) of the cow.
Bessie must choose which cows she wants to bring to her exhibition. She believes that the total smartness TS of the group is the sum of the Si's and, likewise, the total funness TF of the group is the sum of the Fi's. Bessie wants to maximize the sum of TS and TF, but she also wants both of these values to be non-negative (since she must also show that the cows are well-rounded; a negative TS or TF would ruin this). Help Bessie maximize the sum of TS and TF without letting either of these values become negative.
Input
* Line 1: A single integer N, the number of cows
* Lines 2..N+1: Two space-separated integers Si and Fi, respectively the smartness and funness for each cow.
* Lines 2..N+1: Two space-separated integers Si and Fi, respectively the smartness and funness for each cow.
Output
* Line 1: One integer: the optimal sum of TS and TF such that both TS and TF are non-negative. If no subset of the cows has non-negative TS and non- negative TF, print 0.
Sample Input
5-5 78 -66 -32 1-8 -5
Sample Output
8
Hint
OUTPUT DETAILS:
Bessie chooses cows 1, 3, and 4, giving values of TS = -5+6+2 = 3 and TF
= 7-3+1 = 5, so 3+5 = 8. Note that adding cow 2 would improve the value
of TS+TF to 10, but the new value of TF would be negative, so it is not
allowed.
Bessie chooses cows 1, 3, and 4, giving values of TS = -5+6+2 = 3 and TF
= 7-3+1 = 5, so 3+5 = 8. Note that adding cow 2 would improve the value
of TS+TF to 10, but the new value of TF would be negative, so it is not
allowed.
给出每头牛的s和f 要挑出几头牛
要求ts和tf的和最大
切ts和tf不能为负
dp[i]代表ts-100000为i时tf的最大值
因为有负数 所以以100000为起点 i<100000时代表ts<0
#include<cstdio>#include<cstring>#define INF 0x3f3f3f3fint maxn(int a,int b){ return a>b?a:b;}int dp[211111];int main(){ int n,i,j,s[110],f[110]; while(scanf("%d",&n)!=EOF) { for(i=0;i<n;i++) scanf("%d %d",&s[i],&f[i]); for(i=0;i<=200000;i++) dp[i]=-INF; dp[100000]=0; for(i=0;i<n;i++) { if(s[i]>=0) { for(j=200000;j>=s[i];j--) { dp[j]=maxn(dp[j],dp[j-s[i]]+f[i]); } } else { for(j=s[i];j<=200000+s[i];j++) { dp[j]=maxn(dp[j],dp[j-s[i]]+f[i]); } } } int ans=0; for(i=100000;i<=200000;i++) { if(dp[i]>0) ans=maxn(ans,i-100000+dp[i]); } printf("%d\n",ans); } return 0;}
0 0
- POJ 2184 Cow Exhibition ( 01背包变形 )
- poj 2184 Cow Exhibition(01背包变形)
- POJ 2184 Cow Exhibition(01背包变形)
- poj 2184 Cow Exhibition 01背包变形
- POJ-2184-01背包变形-Cow Exhibition
- POJ 2184Cow Exhibition(01背包变形)
- POJ-2184 Cow Exhibition(01背包变形)
- Poj 2184 Cow Exhibition【01背包 变形】
- POJ 2184 Cow Exhibition 01 背包变形
- poj 2184 Cow Exhibition 01背包变形,正负背包
- POJ 2184 Cow Exhibition 01背包的变形
- poj 2184 Cow Exhibition (变形的01背包)
- poj(2184)——Cow Exhibition(01背包变形)
- POJ 2184 Cow Exhibition (01背包变形 + 技巧 好题)
- POJ 2184 Cow Exhibition(01背包变形 or dfs+剪枝)
- POJ 2184 Cow Exhibition(变形01背包)
- POJ 2184 Cow Exhibition(01背包变形)
- NYOJ 2184 Cow Exhibition(01背包+变形)
- DevExpress根据横向扩展表格生成打印表格
- UIView中的坐标转换
- Ubuntu下Sublime Text 2的安装
- 一个简单的ACE Reactor框架的使用
- php小技巧
- poj 2184 Cow Exhibition 01背包变形
- Oracle Forms Services Architecture
- 自定义view中xml属性 提示Attribute "XXX" has already been defined
- ls/cd/pwd/whoami/mkdir/rmdir/rm/pos/top/touch/ln/mv/clear/tar (linux)!!!
- 关于Block的copy和循环引用的问题
- 第九周项目五(三)
- 模板-后缀数组
- codechef The Army
- 单引号,双引号,三引号的区别