HDOJ 5383 Yu-Gi-Oh! 最大費用最大フロー

9825 ワード

ネットワークストリームの裸問題:
2つの部分に分けて図を建てて、満流の最大の費用の最大の流れを要求しないことを求めます.....
Yu-Gi-Oh!
Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 401    Accepted Submission(s): 108
Problem Description
"Yu-Gi-Oh!", also known as "Dueling Monsters", is a popular trading card game which has nearly 20 years history. Next year, YGO will reach its 20th birthday.
Stilwell has 
n  monsters on the desk, each monster has its 
leveli  and 
ATKi . There are two kinds of monsters, Tuner monsters and Non-Tuner monsters.
Now, Stilwell plans to finish some "Synchro Summon", and "Synchro Summon"is a kind of special summon following these rules (a little different from the standard YGO rules):
(1) A "Synchro Summon"needs two monsters as the material of this summon, and they must be one Tuner monster and one Non-Tuner monster.
In other words, we can cost one Tuner monster and one Non-Tuner monster to get a Synchro monster ("cost"means remove form the desk, "get"means put on to the desk).
(2) To simplify this problem, Synchro monsters are neither Tuner monsters nor Non-Tuner monsters.
(3) The level sum of two material must be equal to the level of Synchro monster we summon.
For example:
A Level 3 Tuner monster 
+  A Level 2 Non-Tuner monster 
=  A Level 5 Synchro Monster
A Level 2 Tuner monster 
+  A Level 4 Non-Tuner monster 
=  A Level 6 Synchro Monster
A Level 4 Tuner monster 
+  A Level 4 Non-Tuner monster 
=  A Level 8 Synchro Monster
(4) The material of some Synchro monster has some limits, the material must contain some specific monster.
For example:
A Level 5 Synchro Monster 
α  requires A Level 3 Tuner monster 
α  to be its material
A Level 6 Synchro Monster 
β  requires A Level 4 Non-Tuner monster 
β  to be its material
A Level 8 Synchro Monster 
γ  requires A Level 4 Tuner monster 
γ  
+  A Level 4 Non-Tuner monster 
γ  to be its material
A Level 5 Synchro Monster 
φ  doesn't require any monsters to be its material
Then
A Level 3 Tuner monster 
α  
+  A Level 2 Non-Tuner monster 
=  A Level 5 Synchro Monster 
α
A Level 3 Tuner monster 
δ  
+  A Level 2 Non-Tuner monster 
≠  A Level 5 Synchro Monster 
α
A Level 2 Tuner monster 
+  A Level 4 Non-Tuner monster 
β  
=  A Level 6 Synchro Monster 
β
A Level 3 Tuner monster 
+  A Level 3 Non-Tuner monster 
ζ  
≠  A Level 6 Synchro Monster 
β
A Level 4 Tuner monster 
γ  
+  A Level 4 Non-Tuner monster 
γ  
=  A Level 8 Synchro Monster 
γ
A Level 4 Tuner monster 
σ  
+  A Level 4 Non-Tuner monster 
γ  
≠  A Level 8 Synchro Monster 
γ
A Level 4 Tuner monster 
γ  
+  A Level 4 Non-Tuner monster 
ϕ  
≠  A Level 8 Synchro Monster 
γ
A Level 3 Tuner monster 
+  A Level 2 Non-Tuner monster 
=  A Level 5 Synchro Monster 
φ
A Level 3 Tuner monster 
α  
+  A Level 2 Non-Tuner monster 
=  A Level 5 Synchro Monster 
φ
Stilwell has 
m  kinds of Synchro Monster cards, the quantity of each Synchro Monster cards is infinity.
Now, given 
leveli  and 
ATKi  of every card on desk and every kind of Synchro Monster cards. Please finish some Synchro Summons (maybe zero) to maximum 
∑ATKi  of the cards on desk.
 
Input
The first line of the input contains a single number 
T , the number of test cases.
For each test case, the first line contains two integers 
n , 
m .
Next 
n  lines, each line contains three integers 
tuneri , 
leveli , and 
ATKi , describe a monster on the desk. If this monster is a Tuner monster, then 
tuneri=1 , else 
tuneri=0  for Non-Tuner monster.
Next 
m  lines, each line contains integers 
levelj , 
ATKj , 
rj , and following 
rj  integers are the required material of this Synchro Monster (the integers given are the identifier of the required material).
The input data guarantees that the required material list is available, two Tuner monsters or two Non-Tuner monsters won't be required. If 
ri=2  the level sum of two required material will be equal to the level of Synchro Monster.
T≤10 , 
n,m≤300 , 
1≤leveli≤12 , 
0≤ATKi≤5000 , 
0≤ri≤2
 
Output
T  lines, find the maximum 
∑ATKi  after some Synchro Summons.
 
Sample Input

   
   
   
   
5 2 2 1 3 1300 0 2 900 5 2300 1 1 8 2500 0 2 1 1 3 1300 1 2 900 5 2300 1 1 3 1 1 3 1300 0 2 900 0 2 800 5 2300 1 1 3 1 1 1 233 0 1 233 0 1 200 2 466 2 1 2 6 3 1 3 1300 0 2 900 0 5 1350 1 4 1800 0 10 4000 0 10 1237 5 2300 1 1 8 3000 0 6 2800 0

 
Sample Output

   
   
   
   
2300 2200 3200 666 11037

 
Author
SXYZ
 
/* ***********************************************
Author        :CKboss
Created Time  :2015 08 17      08 42 00 
File Name     :HDOJ5383.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;


const int INF=0x3f3f3f3f;
const int maxv=400;
const int maxn=maxv*maxv;

struct Edge
{
	int to,next,cap,flow,cost;
}edge[maxn];

int n,m;
int Adj[maxv],Size,N;

void init()
{
	memset(Adj,-1,sizeof(Adj)); Size=0;
}

void addedge(int u,int v,int cap,int cost)
{
	edge[Size].to=v;
	edge[Size].next=Adj[u];
	edge[Size].cost=cost;
	edge[Size].cap=cap;
	edge[Size].flow=0;
	Adj[u]=Size++;
}

void Add_Edge(int u,int v,int cap,int cost)
{
	addedge(u,v,cap,cost);
	addedge(v,u,0,-cost);
}

int dist[maxv];
int vis[maxv],pre[maxv];

bool spfa(int s,int t)
{
	queue<int> q;
	for(int i=0;i<N;i++)
	{
		dist[i]=-INF; vis[i]=false; pre[i]=-1;
	}
	dist[s]=0; vis[s]=true; q.push(s);
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		vis[u]=false;
		for(int i=Adj[u];~i;i=edge[i].next)
		{
			int v=edge[i].to;
			if(edge[i].cap>edge[i].flow&&
					dist[v]<dist[u]+edge[i].cost)
			{
				dist[v]=dist[u]+edge[i].cost;
				pre[v]=i;
				if(!vis[v])
				{
					vis[v]=true;
					q.push(v);
				}
			}
		}
	}
	if(pre[t]==-1) return false;
	return true;
}

int MinCostMaxFlow(int s,int t,int &cost)
{
	int flow=0;
	cost=0;
	while(spfa(s,t))
	{
		int Min=INF;
		for(int i=pre[t];~i;i=pre[edge[i^1].to])
		{
			if(Min>edge[i].cap-edge[i].flow)
				Min=edge[i].cap-edge[i].flow;
		}
		if(dist[t]<0) break;
		for(int i=pre[t];~i;i=pre[edge[i^1].to])
		{
			edge[i].flow+=Min;
			edge[i^1].flow-=Min;
			cost+=edge[i].cost*Min;
		}
		flow+=Min;
	}
	return flow;
}

struct Moster
{
	Moster(){}
	Moster(int l,int a):level(l),ATK(a){}
	int level,ATK;
};

vector<Moster> m0,m1;
int turn[maxv],pos[maxv];
int GG[maxv][maxv];

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%d%d",&n,&m);
		init(); m0.clear(); m1.clear();
		memset(GG,0,sizeof(GG));
		int sumATK=0;
		int sz1=0,sz2=0;
		for(int i=0,t,l,a;i<n;i++)
		{
			scanf("%d%d%d",&t,&l,&a);
			if(t==0) 
			{
				m0.push_back(Moster(l,a));
				turn[i]=0; pos[i]=sz1++;
			}
			else if(t==1) 
			{
				m1.push_back(Moster(l,a));
				turn[i]=1; pos[i]=sz2++;
			}
			sumATK+=a;
		}
		for(int i=0,l,a,r;i<m;i++)
		{
			scanf("%d%d%d",&l,&a,&r);
			if(r==0)
			{
				for(int j=0;j<sz1;j++)
				{
					for(int k=0;k<sz2;k++)
					{
						int u=j+1,v=k+sz1+1;
						if(m0[j].level+m1[k].level==l)
						{
							if(a>m0[j].ATK+m1[k].ATK)
							{
								GG[u][v]=max(GG[u][v],a-m0[j].ATK-m1[k].ATK);
							}
						}
					}
				}
			}
			else if(r==1)
			{
				int x;
				scanf("%d",&x); x--;
				if(turn[x]==0)
				{
					int P=pos[x];
					for(int j=0;j<sz2;j++)
					{
						int u=P+1,v=j+sz1+1;
						if(m0[P].level+m1[j].level==l)
						{
							if(a>m0[P].ATK+m1[j].ATK)
							{
								GG[u][v]=max(GG[u][v],a-m0[P].ATK-m1[j].ATK);
							}
						}
					}
				}
				else if(turn[x]==1)
				{
					int P=pos[x];
					for(int j=0;j<sz1;j++)
					{
						int u=j+1,v=P+sz1+1;
						if(m0[j].level+m1[P].level==l)
						{
							if(a>m0[j].ATK+m1[P].ATK)
							{
								GG[u][v]=max(GG[u][v],a-m0[j].ATK-m1[P].ATK);
							}
						}
					}
				}
			}
			else if(r==2)
			{
				int x,y;
				scanf("%d%d",&x,&y); x--; y--;
				if(turn[x]==1) swap(x,y);
				int u=pos[x]+1,v=sz1+pos[y]+1;
				if(a>m0[pos[x]].ATK+m1[pos[y]].ATK)
				{
					GG[u][v]=max(GG[u][v],a-m0[pos[x]].ATK-m1[pos[y]].ATK);
				}
			}
		}

		for(int i=1;i<=sz1;i++)
		{
			for(int j=sz1+1;j<=sz1+sz2;j++)
			{
				if(GG[i][j]>0) Add_Edge(i,j,1,GG[i][j]);
			}
		}

		int S=0,T=sz1+sz2+1;
		for(int i=1;i<=sz1;i++) Add_Edge(0,i,1,0);
		for(int i=sz1+1;i<=sz1+sz2;i++) Add_Edge(i,T,1,0);

		int flow,cost; N=sz1+sz2+2;
		flow=MinCostMaxFlow(S,T,cost);
		printf("%d
",sumATK+cost); } return 0; }