Mr. Panda and Crystal
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 979 Accepted Submission(s): 322
Problem Description
Long long time ago, there is a magic continent far far away.
There are N types of magic crystals that contain ancient magic powers. Each of the type of magic crystal has its own price for one piece in the market. As the most powerful magician, Mr. Panda could synthesize some types of crystals by collecting some amount of other types of crystals. He could also create some types of crystals by using some number of his magic powers.
Now, Mr Panda can create any number of crystals as he wish by using no more than M magic powers. He want to know the maximum amount of money he can make by sell all the crytals he creates and synthesizes.
Input
The first line of the input gives the number of test cases, T. T test cases follow.
Each test case starts with 3 positive intergers, M, N and K represent the amount of magic powers Mr. Panda had, the number of crystal types on the magic continent and the number of crystal synthesis equations.
Then N lines follows, each of them starts with one 0 or 1 which indicates whehter Mr. Panda could create this type of crystal.
If the ith line starts with 0, which means Mr. Panda couldn’t create crystal type i. Then there is one integer pi in this line which is the price for each piece of crystal type i.
If the ith line starts with 1, which means Mr. Panda could create crystal type i. Then there are two positive integers ci and pi in this line, the first is the amout of magic power cost when creates one piece of crystal type i, and the second is is the price for each piece of crystal type i.
The following K lines each start with two interger xi and yi , which means for synthesizing one piece of crystal type xi , yi rules should be satisfied. Then there are yipair of positive intergers uj and vj means for one piece of xthi type cristal, we have to collect vi piece of crystal type ui. Only when all the rules of ui and vi are satisfied, Mr. Panda could synthesize one piece xthi type cristal.
Output
For each test case, output one line containing “Case #x: y”, where x is the test case number (starting from 1) and y is the maximum amout of money Mr. Panda could make.
limits
∙1≤T≤100.
∙1≤M≤10000.
∙1≤N≤200.
∙1≤K≤200.
∙1≤xi,uj≤N.
∙foreachcrystalsynthesisequation,allujaredifferent.
∙1≤vj≤100.
∙1≤ci,pi≤10000.
Sample Input
2 100 3 2 0 20 1 15 10 1 2 1 1 2 2 1 3 1 2 1 3 2 100 3 2 1 3 1 1 4 1 0 10 3 1 1 3 3 1 2 2
Sample Output
Case #1: 330 Case #2: 121
題意：
給你n(n<=1e4),m(m<=2e2),k(k<=2e2)。你有n個魔法值。
接下來m行給出m種魔法石，每行格式爲 id 如果id爲0則爲 id x 表示這個魔法石不可以合成，x即爲賣出的價格。
如果id爲1，則爲 id x y，表示可以用x的魔法值合成這塊魔法石，賣出可以得到y元。
接下來k行，給出k種合成方法。
每一行格式爲 id x 表示第id種魔法石可以由x種其他魔法石合成。再給你2*x個數，每一對數<y,z>表示需要z塊第y種魔法石。
問你n個魔法值最多可以賺多少錢。
題解：
很容易想到當每一種魔法石的合成最小消耗魔法值確定時，就是一個容量爲n的完全揹包。
但是最關鍵的是怎麼求每一種魔法石的合成的最小消耗。
於是可以想到，利用dijkstra算法求最短路的思想，每一種魔法石向包含他的合成方案編號建邊，將距離改爲合成該方案的魔法石所消耗的總魔力值。如果對dijk算法非常熟練的話，那麼這道題就很簡單了。注意無法合成的魔法石重量置爲n+1，否則可能會RE。
主要是還是要讀懂題目。
代碼：
#include<bits/stdc++.h>
#define ll long long
#define inf 100000000000000000LL
using namespace std;
const int maxn=10010;
int n,m,k;
struct node
{
int id;
int dis;
node(int x=0,int y=0){id=x;dis=y;}
bool operator<(node aa)
const{
return dis>aa.dis;
}
};
struct wupin
{
int xh;
int mon;
}a[maxn];
struct edge
{
int he;
vector<pair<int,int> >vc;
}p[210];
priority_queue<node>q;
int d[maxn];
bool vis[maxn];
vector<int>vv[maxn];
int getsum(edge &e)
{
int sum=0;
for(int i=0;i<e.vc.size();i++)
{
sum+=e.vc[i].second*d[e.vc[i].first];
}
return sum;
}
void dijk()
{
while(!q.empty())
{
node no=q.top();q.pop();
int u=no.id;
if(vis[u]) continue;
vis[u]=1;
for(int i=0;i<vv[u].size();i++)
{
edge &e=p[vv[u][i]];
int tmp=getsum(e);
if(d[e.he]>tmp)
{
d[e.he]=tmp;
q.push(node(e.he,d[e.he]));
}
}
}
}
int dp[maxn];
int main()
{
int T,cas=1;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&k);
while(!q.empty()) q.pop();
memset(vis,0,sizeof(vis));
for(int i=1;i<=m;i++)
{
vv[i].clear();
int x,y;
scanf("%d",&x);
if(x==1)
{
scanf("%d%d",&x,&y);
a[i].xh=x;
a[i].mon=y;
d[i]=x;
}
else
{
scanf("%d",&y);
a[i].mon=y;
a[i].xh=n+1;
d[i]=n+1;
}
}
for(int i=0;i<k;i++)
{
p[i].vc.clear();
scanf("%d",&p[i].he);
int x,y,z;
scanf("%d",&x);
for(int j=0;j<x;j++)
{
scanf("%d%d",&y,&z);
p[i].vc.push_back(make_pair(y,z));
vv[y].push_back(i);
}
}
for(int i=1;i<=m;i++)
if(d[i]<=n){
q.push(node(i,d[i]));
}
dijk();
memset(dp,0,sizeof(dp));
for(int i=1;i<=m;i++)
for(int j=d[i];j<=n;j++)
{
dp[j]=max(dp[j],dp[j-d[i]]+a[i].mon);
}
printf("Case #%d: %d\n",cas++,dp[n]);
}
return 0;
}
/*
2
100 3 2
0 20
1 15 10
1 2 1
1 2 2 1 3 1
2 1 3 2
100 3 2
1 3 1
1 4 1
0 10
3 1 1 3
3 1 2 2
*/