ZOJ 3905 Cake

Cake

---

Time Limit: 4 Seconds      Memory Limit: 65536 KB

---

Alice and Bob like eating cake very much. One day, Alice and Bob went to a bakery and bought many cakes.

Now we know that they have bought n cakes in the bakery. Both of them like delicious cakes, but they evaluate the cakes as different values. So they decided to divide those cakes by following method.

Alice and Bob do n / 2 steps, at each step, Alice choose 2 cakes, and Bob takes the cake that he evaluates it greater, and Alice take the rest cake.

Now Alice want to know the maximum sum of the value that she can get.

Input

The first line is an integer T which is the number of test cases.

For each test case, the first line is an integer n (1<=n<=800). Note that n is always an even integer.

In following n lines, each line contains two integers a[i] and b[i], where a[i] is the value of i^th cake that Alice evaluates, and b[i] is the value of i^th cake that Bob evaluates. (1<=a[i], b[i]<=1000000)

Note that a[1], a[2]..., a[n] are n distinct integers and b[1], b[2]..., b[n] are n distinct integers.

Output

For each test case, you need to output the maximum sum of the value that Alice can get in a line.

Sample Input

161 67 106 1112 1815 52 14

Sample Output

28

先按b从大到小排序，然后像做背包一样搞就好了

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=1005;int T,n,m,f[maxn][maxn];struct point{int a,b;void read(){scanf("%d%d",&a,&b);}bool operator<(const point&f) const{return b>f.b;}}u[maxn];int main(){scanf("%d",&T);while (T--){memset(f,0,sizeof(f));scanf("%d",&n);for (int i=1;i<=n;i++) u[i].read();sort(u+1,u+n+1);for (int i=1;i<=n;i++){f[i][0]=0;for (int j=1;j+j<=i;j++){f[i][j]=max(f[i-1][j],f[i-1][j-1]+u[i].a);}}printf("%d\n",f[n][n/2]);}return 0;}