题意：你生活在一个魔法大陆上，你有n 魔力， 这个大陆上有m 种魔法水晶，还有n 种合成水晶的方式，每种水晶价格告诉你，并且告诉你哪些水晶你能直接造出来，哪些你必须合成才能造出来，问你n魔力最多能卖多少钱的水晶？
析：首先知道的是，如果每个所消耗的魔法水晶固定，那么这就是一个背包问题，很简单就能搞定，然而并不是，但是我们能够推出，经过一定次数的变换，这个值肯定就是固定的了而且是最小的，这个就可以用spfa进行松弛操作。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e15;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 200 + 50;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
int val[maxn], w[maxn];
int num[maxn];
int dp[maxn*50];
vector<P> v[maxn];
void solve(int u){
int sum = 0;
for(int i = 0; i < v[u].sz; ++i)
sum += w[v[u][i].fi] * v[u][i].se;
if(sum < w[num[u]]) w[num[u]] = sum;
}
int main(){
int T; cin >> T;
for(int kase = 1; kase <= T; ++kase){
int k;
scanf("%d %d %d", &m, &n, &k);
for(int i = 1; i <= n; ++i){
int x;
scanf("%d", &x);
if(0 == x) scanf("%d", val + i), w[i] = m + 1;
else scanf("%d %d", w+i, val+i);
}
for(int i = 1; i <= k; ++i){
v[i].cl;
scanf("%d", num+i);
int x;
scanf("%d", &x);
for(int j = 0; j < x; ++j){
int u, vv;
scanf("%d %d", &u, &vv);
v[i].pb(P(u, vv));
}
}
for(int i = 1; i <= 5; ++i){
for(int j = 1; j <= k; ++j)
solve(j);
}
ms(dp, 0);
for(int i = 1; i <= n; ++i)
for(int j = w[i]; j <= m; ++j)
dp[j] = max(dp[j], dp[j-w[i]] + val[i]);
printf("Case #%d: %d
", kase, dp[m]);
}
return 0;
}