2014
02-12

# Accepted Necklace

I have N precious stones, and plan to use K of them to make a necklace for my mother, but she won’t accept a necklace which is too heavy. Given the value and the weight of each precious stone, please help me find out the most valuable necklace my mother will accept.

The first line of input is the number of cases.
For each case, the first line contains two integers N (N <= 20), the total number of stones, and K (K <= N), the exact number of stones to make a necklace.
Then N lines follow, each containing two integers: a (a<=1000), representing the value of each precious stone, and b (b<=1000), its weight.
The last line of each case contains an integer W, the maximum weight my mother will accept, W <= 1000.

The first line of input is the number of cases.
For each case, the first line contains two integers N (N <= 20), the total number of stones, and K (K <= N), the exact number of stones to make a necklace.
Then N lines follow, each containing two integers: a (a<=1000), representing the value of each precious stone, and b (b<=1000), its weight.
The last line of each case contains an integer W, the maximum weight my mother will accept, W <= 1000.

1
2 1
1 1
1 1
3 

1

#include <stdio.h>
#include <string.h>
struct node {
int v,w;

} a[21];
#define INF 0xfffffff
int f[1005][21];
int n, k,w;
void dp() {
int i,j, p;
int max;
memset(f,0,sizeof(f));
for(i=0; i<n; i++) {
for(j=w; j>=a[i].w; j--) {
for(p=1; p<=k; p++) {
max = f[j][p];
if(max<f[j-a[i].w][p-1]+a[i].v) max = f[j-a[i].w][p-1]+a[i].v;
f[j][p] = max;
}
}

}
max =-INF;
for(i=0; i<w; i++) if(max<f[i][k]) max = f[i][k];
printf("%d\n",max);
}
int main() {
int T,i;
scanf("%d",&T);
while(T--) {
scanf("%d%d",&n,&k);
for(i=0; i<n; i++) scanf("%d%d",&a[i].v,&a[i].w);
scanf("%d",&w);
dp();
}
return 0;
}

1. 题目需要求解的是最小值，而且没有考虑可能存在环，比如
0 0 0 0 0
1 1 1 1 0
1 0 0 0 0
1 0 1 0 1
1 0 0 0 0
会陷入死循环