HDU 5855 Less Time, More profit 最大权闭合子图

题目链接：

Less Time, More profit

问题描述

输入

输出

样例

题意

题解

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
//#define lson (o<<1)
//#define rson ((o<<1)|1)
//#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=444;

struct Edge {
int u,v,cap,flow;
Edge(int u,int v,int c,int f):u(u),v(v),cap(c),flow(f) {}
};

struct Dinic {
int n,m,s,t;
vector<Edge> egs;
vector<int> G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn];

void init(int n) {
this->n=n;
rep(i,0,n+1) G[i].clear();
egs.clear();
m=0;
}

void addEdge(int u,int v,int cap) {
egs.pb(Edge(u,v,cap,0));
egs.pb(Edge(v,u,0,0));
m=egs.sz();
G[u].pb(m-2);
G[v].pb(m-1);
}

bool BFS() {
clr(vis,0);
queue<int> Q;
Q.push(s);
d[s]=0;
vis[s]=1;
while(!Q.empty()) {
int x=Q.front();
Q.pop();
rep(i,0,G[x].sz()) {
Edge& e=egs[G[x][i]];
if(!vis[e.v]&&e.cap>e.flow) {
vis[e.v]=1;
d[e.v]=d[x]+1;
Q.push(e.v);
}
}
}
return vis[t];
}

int DFS(int x,int a) {
if(x==t||a==0) return a;
int flow=0,f;
for(int& i=cur[x]; i<G[x].size(); i++) {
Edge& e=egs[G[x][i]];
if(d[x]+1==d[e.v]&&(f=DFS(e.v,min(a,e.cap-e.flow)))>0) {
e.flow+=f;
egs[G[x][i]^1].flow-=f;
flow+=f;
a-=f;
if(a==0) break;
}
}
return flow;
}

int Maxflow(int s,int t) {
this->s=s;
this->t=t;
int flow=0;
while(BFS()) {
clr(cur,0);
flow+=DFS(s,INF);
}
return flow;
}
} dinic;

PII plants[maxn];
int pi[maxn],vis[maxn];
vector<int> G[maxn];

int n,m,L;

int isok(int mid) {
dinic.init(n+m+2);
clr(vis,0);
for(int i=1; i<=m; i++) {
rep(j,0,G[i].size()) {
int v=G[i][j];
if(plants[v].Y>mid) {
vis[i]=1;
break;
}
}
}
int sumv=0;
for(int i=1; i<=m; i++) {
if(vis[i]) continue;
sumv+=pi[i];
dinic.addEdge(0,i,pi[i]);
rep(j,0,G[i].size()) {
int v=G[i][j];
dinic.addEdge(i,v+m,INF);
}
}
rep(i,1,n+1) {
if(plants[i].Y>mid) continue;
dinic.addEdge(i+m,m+n+1,plants[i].X);
}
return sumv-dinic.Maxflow(0,n+m+1);
}

void init() {
rep(i,0,maxn) G[i].clear();
}

int main() {
int tc,kase=0;
scanf("%d",&tc);
while(tc--) {
scanf("%d%d%d",&n,&m,&L);
init();
rep(i,1,n+1) {
scanf("%d%d",&plants[i].X,&plants[i].Y);
}
rep(i,1,m+1) {
scanf("%d",&pi[i]);
int cnt;
scanf("%d",&cnt);
while(cnt--) {
int x;
scanf("%d",&x);
G[i].push_back(x);
}
}
printf("Case #%d: ",++kase);
int l=0,r=1000000000;
if(isok(r)<L) {
puts("impossible");
continue;
}
while(l+1<r) {
int mid=l+(r-l)/2;
if(isok(mid)>=L) r=mid;
else l=mid;
}
printf("%d %d\n",r,isok(r));
}
return 0;
}

//end-----------------------------------------------------------------------
/*
5 3 50
1 1
2 2
3 3
4 4
5 5
100 2 1 4
1 0
1 0
*/