HLG 1053 Warcraft III（背包DP）

Warcraft III

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Time Limit: 10000 MSMemory Limit: 65536 KTotal Submit: 458(136 users)Total Accepted: 181(120 users)Rating:Special Judge: NoDescription

dccmx likes playing Warcraft III. Now, he is teaching his girlfriend to play it. In Warcraft III, there are many kinds of units. Every unit costs some gold and lumber. Different units have different attack value.

Now question comes. Given some amount of gold and a list of types of units, how to arrange your units to maximize the attack value of your units. Assume you have infinite lumbers.

Input

Line 1 contains an integer T: the number of test cases.

Next T blocks, each starts with two integers: G and U, represents the amount of gold and number of unit type. Next U lines, each contains two integers: the attack value of a type of unit and the cost.

Output

For each test case, output the maximum total attack value in one line.

Sample Input

2

100 1

20 10

300 4

100 60

250 120

120 100

35 20

Sample Output

200

605

裸的完全背包问题；

代码：

#include <iostream>#include<string.h>#include<stdio.h>#define N 10000#define max(a,b) a>b?a:busing namespace std;int main(){    int t;    int value[N],dp[N],v[N],V,n,i,j;    while(~scanf("%d",&t))    {        while(t--)        {            memset(dp,0,sizeof(dp));            scanf("%d%d",&V,&n);            for(i=0;i<n;i++)            {                scanf("%d%d",&value[i],&v[i]);            }            for(i=0;i<n;i++)            {                for(j=v[i];j<=V;j++)                {                    dp[j]=max(dp[j],dp[j-v[i]]+value[i]);                }            }            cout<<dp[V]<<endl;        }    }    return 0;}