HDU 6007 Mr. Panda and Crystal (最短路 + 完全背包)

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？

思路：

很复杂的一道题目。

但是分析后，很简单：

整体思路：

我们求出每种水晶的最小消耗魔力，那么问题就转换为 求在不超过魔力n 的情况下，让你选择尽可能多水晶来卖更多的钱。显然，是一个完全背包。

那么问题又进一步转换为求每种水晶的最小消耗魔力。

因为合成是能传递的。

所以我们按照类似dijkstra的方式写就行，把处理最小的魔力消耗 水晶，来更新能合成的水晶的最小魔力消耗。

分析完是不是觉的很水。=_=~

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;

int T, n, m, k, ks;

const int inf = 0x3f3f3f3f;
const int maxn = 10000 + 10;
const int maxm = 200 + 10;

struct Thing{
int money;
int xh;
}tg[maxn];

struct Edge{
int goal;
vector<pair<int,int> >sub;
}p[maxm];

int dp[maxn];

struct Node{
int u,d;
Node(int u = 0,int d = 0):u(u),d(d){}

bool operator < (const Node& rhs) const {
return d > rhs.d;
}
};

vector<int>g[maxm];
int d[maxm];
bool vis[maxm];
priority_queue<Node>q;

int get_sum(Edge& e){
int sum = 0;
for (int i = 0; i < e.sub.size(); ++i){
sum += e.sub[i].second * d[e.sub[i].first];
}
return sum;
}

void dij(){
while(!q.empty()){
Node nod = q.top(); q.pop();
int u = nod.u, dis = nod.d;
if (vis[u]) continue;
vis[u] = 1;
for (int i = 0; i < g[u].size(); ++i){
Edge& e = p[g[u][i]];
int tmp = get_sum(e);
if (d[e.goal ] > tmp ){
d[e.goal] = tmp;
q.push(Node(e.goal, d[e.goal]));
}
}
}
}

int main(){
scanf("%d", &T);
while(T--){
memset(vis,0,sizeof vis);
while(!q.empty())q.pop();
scanf("%d %d %d",&n, &m, &k);

for (int i = 1; i <= m; ++i){
g[i].clear();
Thing& t = tg[i];
int x,y;
scanf("%d",&x);
if (x == 0){
t.xh = n+1;
d[i] = n+1;
scanf("%d",&t.money);
}
else{
scanf("%d%d",&y,&t.money);
t.xh = y;
d[i] = y;
}
}
for (int i = 0; i < k; ++i){
Edge& e = p[i];
e.sub.clear();
scanf("%d",&e.goal);
int num;
scanf("%d",&num);
for (int j = 0; j < num; ++j){
int x,y;
scanf("%d%d",&x, &y);
e.sub.push_back(make_pair(x,y));
g[x].push_back(i);
}
}

for (int i = 1; i <= m; ++i){
if (d[i] <= n){
q.push(Node(i, d[i]));
}
}

dij();

memset(dp,0,sizeof dp);
for (int i = 1; i <= m; ++i){
for (int j = 0; j <= n; ++j){
if (j-d[i] >= 0) dp[j] = max(dp[j],dp[j-d[i] ] + tg[i].money);
}
}
printf("Case #%d: %d\n",++ks, dp
);
}

return 0;
}

Mr. Panda and Crystal

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 22    Accepted Submission(s): 12


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 yi pair
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

 

Source

2016 CCPC-Final