说是最大权闭合图....

比赛时没敢写....

题意

一共有n个任务，m个技术

完成一个任务可盈利一些钱，学习一个技术要花费钱

完成某个任务前需要先学习某几个技术

但是可能在学习一个任务前需要学习另几个任务

求最多能赚多少钱咯

先将缩点将需要一起学掉的技术缩成一个点

建s--任务 权值为该任务盈利多少钱

建技术（缩点后）-t 权值为学习这技术的花费（总）

任务-技术 （完成该任务所需的每个技术都需要建边）权值为INF

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<limits.h>
#include<ctype.h>
#include<math.h>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
#include<queue>
#include<stack>
#include<vector>
#include<deque>
#include<set>
#include<map>
//N为最大点数
#define M 150
//M为最大边数
const int MAXN = 2999;//点数的最大值
const int MAXM = 2222;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge{int from, to, nex;bool sign;//是否为桥
}edge[M<<1];
int head[MAXN], edgenum;
void add(int u, int v){//边的起点和终点Edge E={u, v, head[u], false};edge[edgenum] = E;head[u] = edgenum++;
}int DFN[333], Low[333], Stack[333], top, Time; //Low[u]是点集{u点及以u点为根的子树} 中(所有反向弧)能指向的(离根最近的祖先v) 的DFN[v]值(即v点时间戳)
int taj;//连通分支标号，从1开始
int Belong[333];//Belong[i] 表示i点属于的连通分支
bool Instack[333];
vector<int> bcc[333]; //标号从1开始void tarjan(int u ,int fa)
{DFN[u] = Low[u] = ++ Time ;Stack[top ++ ] = u ;Instack[u] = 1 ;for (int i = head[u] ; ~i ; i = edge[i].nex ){int v = edge[i].to;if(DFN[v] == -1){tarjan(v , u);Low[u] = min(Low[u] ,Low[v]) ;if(DFN[u] < Low[v]){edge[i].sign = 1;//为割桥}}else if(Instack[v]){Low[u] = min(Low[u] ,DFN[v]) ;}}if(Low[u] == DFN[u]){int now;taj ++ ;bcc[taj].clear();do{now = Stack[-- top] ;Instack[now] = 0 ;Belong [now] = taj ;bcc[taj].push_back(now);}while(now != u) ;}
}void tarjan_init(int all){memset(DFN, -1, sizeof(DFN));memset(Instack, 0, sizeof(Instack));top = Time = taj = 0;for(int i=1;i<=all;i++)if(DFN[i]==-1 )tarjan(i, i); //注意开始点标！！！
}
vector<int>G[333];
int du[333];
void suodian(){memset(du, 0, sizeof(du));for(int i = 1; i <= taj; i++)G[i].clear();for(int i = 0; i < edgenum; i++){int u = Belong[edge[i].from], v = Belong[edge[i].to];if(u!=v){G[u].push_back(v), du[v]++;}}
}
struct Edge1
{int to,next,cap,flow;
}edge1[MAXM];//注意是MAXM
int tol;
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
//加边，单向图三个参数，双向图四个参数
void addedge (int u,int v,int w,int rw=0)
{edge1[tol].to = v;edge1[tol].cap = w;edge1[tol].next = head[u];edge1[tol].flow = 0;head[u] = tol++;edge1[tol].to = u;edge1[tol].cap = rw;edge1[tol]. next = head[v];edge1[tol].flow = 0;head[v]=tol++;
}
//输入参数：起点、终点、点的总数
//点的编号没有影响，只要输入点的总数
int sap(int start,int end, int N)
{memset(gap,0,sizeof(gap));memset(dep,0,sizeof(dep));memcpy(cur,head,sizeof(head));int u = start;pre[u] = -1;gap[0] = N;int ans = 0;int i;while(dep[start] < N){if(u == end){int Min = INF;for( i = pre[u];i != -1; i = pre[edge1[i^1]. to]){if(Min > edge1[i].cap - edge1[i]. flow)Min = edge1[i].cap - edge1[i].flow;}for( i = pre[u];i != -1; i = pre[edge1[i^1]. to]){edge1[i].flow += Min;edge1[i^1].flow -= Min;}u = start;ans += Min;continue;}bool flag =  false;int v;for( i = cur[u]; i != -1;i = edge1[i].next){v = edge1[i]. to;if(edge1[i].cap - edge1[i].flow && dep[v]+1 == dep[u]){flag =  true;cur[u] = pre[v] = i;break;}}if(flag){u = v;continue;}int Min = N;for( i = head[u]; i !=  -1;i = edge1[i]. next){if(edge1[i].cap - edge1[i].flow && dep[edge1[i].to] < Min){Min = dep[edge1[i].to];cur[u] = i;}}gap[dep[u]]--;if(!gap[dep[u]]) return ans;dep[u] = Min+1;gap[dep[u]]++;if(u != start) u = edge1[pre[u]^1].to;}return ans;
}
int a[MAXN],b[MAXN],cost[MAXN];
vector<int>q[222];
void init(){memset(head, -1, sizeof(head)); edgenum=0;tol=0;}
int main()
{int t,cas=1;int n, m;//n m 为点数和边数//  freopen("in.txt","r",stdin);scanf("%d",&t);while(t--){init();int sum=0;scanf("%d%d",&n,&m);for(int i=1;i<=n;i++){scanf("%d",&a[i]);sum+=a[i];}for(int i=1;i<=m;i++){cost[i]=0;scanf("%d",&b[i]);}for(int i=1;i<=n;i++){int k,c;q[i].clear();scanf("%d",&k);for(int j=0;j<k;j++){scanf("%d",&c);q[i].push_back(c);}}for(int i=1;i<=m;i++)for(int j=1;j<=m;j++){int a;scanf("%d",&a);if(a)add(i,j);}tarjan_init(m);suodian();for(int i=1;i<=m;i++){int v=Belong[i];cost[v]+=b[i];}init();int s=0,end=taj+n+1;for(int i=1;i<=n;i++){addedge(s,taj+i,a[i],0);}for(int i=1;i<=n;i++){for(int j=0;j<q[i].size();j++){addedge(taj+i,Belong[q[i][j]+1],INF,0);}}for(int i=1;i<=taj;i++){addedge(i,end,cost[i],0);for(int j=0;j<G[i].size();j++){addedge(i,G[i][j],INF,0);}}printf("Case #%d: ",cas++);printf("%d\n",sum-sap(s,end,taj+n+2));}return 0;
}