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codevs1823 工作安排

2015-01-18 10:29 134 查看
源向每个工作连流量c[i]的边,费用0

每个人向汇点连s[i]+1条边

流量t[i][j]-t[i][j-1],费用w[i][j]

工作和人根据A矩阵互相连边

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<queue>
#define R0(i,n) for(int i=0;i<n;++i)
#define R1(i,n) for(int i=1;i<=n;++i)
#define cl(x,c) memset(x,c,sizeof x)
#define maxn 1000010
#define INF 707406378
using namespace std;
typedef long long ll;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
struct Edge {int from, to, cap, flow, cost;};
int S, T;ll ans=0;
vector<int> G[maxn];
vector<Edge> edges;
void add(int from, int to, int cap, int cost) {
	edges.push_back((Edge){from, to, cap, 0, cost});
	edges.push_back((Edge){to, from, 0, 0, -cost});
	int m = edges.size();
	G[from].push_back(m-2);
	G[to].push_back(m-1);
}
int d[maxn], p[maxn], A[maxn];
bool inq[maxn];
int cnt=1;
int c[251],a[251][251],s[251],t[251][11],w[251][11];
bool bf() {
	cl(inq,0);
	for(int i = S; i <= T; i++) d[i] = INF;
	d[S] = 0; inq[S] = 1; p[S] = 0; A[S] = INF;
	queue<int> Q;
	Q.push(S);
	while(!Q.empty()) {
		int x = Q.front(); Q.pop();
		inq[x] = 0;
		R0(i,G[x].size()){
			Edge& e = edges[G[x][i]];
			if(e.cap > e.flow && d[e.to] > d[x] + e.cost) {
				d[e.to] = d[x] + e.cost;
				A[e.to] = min(A[x], e.cap-e.flow);
				p[e.to] = G[x][i];
				if(!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }
			}
		}
	}
	if(d[T] == INF) return 0;
	
	ans += d[T]*A[T];
	int x = T;
	while(x != S) {
		edges[p[x]].flow += A[T];
		edges[p[x]^1].flow -= A[T];
		x = edges[p[x]].from;
	}
	
	return 1;
}

int main() {
	S=1,T=1001;
	int m=read(),n=read();
	R1(i,n)c[i]=read();
	R1(i,m)R1(j,n)a[i][j]=read();
	R1(i,m){
		s[i]=read();
		R1(j,s[i])t[i][j]=read();
		t[i][s[i]+1]=INF;
		R1(j,s[i]+1)
		w[i][j]=read();
	}
	R1(i,n)add(0,i,c[i],0);
	R1(i,m)R1(j,n)if(a[i][j])add(j,n+i,INF,0);
	R1(i,m)R1(j,s[i]+1)add(n+i,T,t[i][j]-t[i][j-1],w[i][j]);
	while(bf());
	printf("%lld",ans);
	return 0;
}
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