Like to play Warcraft LTL

Enjoy playing Warcraft's LTL time limit: -Ms | Memory Limit:65535KB Difficulty:3

Describe

LTL is very fond of playing Warcraft, because Warcraft is very fastidious about the overall strength of the team, and he is now also for upgrading without dragging down the team and efforts.

He now has a lot of places to choose to go to the Spawner upgrade, but in every location he had to buy adequate supplies and the right props, in order to avoid monsters when the monster anti-killing, each location of the monster has not been finished (but also do not lose money and equipment), and he just selected the location will be finished brush the site all the monsters , and the corresponding experience values are obtained. Now LTL can give every location the money to buy supplies and props and the experience of hitting all the monsters, but he has a certain amount of money. So he wanted to know how to choose the place that made him the most experienced.

Input

The

first line is an integer T that represents the number of groups of test data 0<t<=10
The second row of two integers n,m,0<n<=100,0<m<=1000000 indicates that the LTL has N different locations of choice and M's total money
The next n rows of two integers per line ci,vi, (0<ci<=1000000,0<vi<=2000) represents the total amount of money and total experience that LTL will need to purchase supplies and props to finish the first I site

.

Output

an integer line representing the maximum XP value that the LTL can obtain

Sample input

23 107 72 33 52 53 52 1

Sample output

Max Experience:12max Experience:6

After reading the question, immediately determine the 01 knapsack problem, and then directly write code, unfortunately, the tle is not time to!

 timeout code: 

 #include <stdio.h>struct node{int c;int W;} Num[105];int dp[1000005];int Max (int a,int b) {return a>b?a:b;} int main () {int t,n,m;int i,j;scanf ("%d", &t), while (t--) {scanf ("%d%d", &n,&m), and for (i=0;i<n;i++) {scanf ( "%d%d", &NUM[I].C,&NUM[I].W);} for (i=0;i<n;i++) {for (j=m;j>=num[i].c;j--) {Dp[j]=max (DP[J],DP[J-NUM[I].C]+NUM[I].W);}} printf ("Max Experience:%d\n", Dp[m]);} return 0;} 
 
 The following is optimized code: 

 #include <stdio.h>struct node{int c;int W;} Num[105];int dp[1000005];int Max (int a,int b) {return a>b?a:b;} int main () {int t,n,m;int i,j,s,count;scanf ("%d", &t), while (t--) {scanf ("%d%d", &n,&m); S=0;for (i=0;i<n ; i++) {scanf ("%d%d", &NUM[I].C,&NUM[I].W); s+=num[i].c;} for (i=0;i<=m;i++) dp[i]=0;for (i=0;i<n;i++) {S-=num[i].c;count=max (M-S,NUM[I].C), for (j=m;j>=count;j--) { Dp[j]=max (DP[J],DP[J-NUM[I].C]+NUM[I].W);}} printf ("Max Experience:%d\n", Dp[m]);} return 0;} 
 
 

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Like to play Warcraft LTL