【BZOJ】【1449】【JSOI2009】球队收益

网络流/费用流/二分图最小权匹配

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　　题解：http://blog.csdn.net/huzecong/article/details/9119741

　　太神了！由于一赢一输不好建图，就先假设全部都输，再将赢的收益修改！就变成普通的二分图了！！

　　费用与流量的平方相关时拆边……这个稍微处理一下即可

/**************************************************************
    Problem: 1449
    User: Tunix
    Language: C++
    Result: Accepted
    Time:676 ms
    Memory:3940 kb
****************************************************************/
 
//BZOJ 1449
#include<cmath>
#include<vector>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define rep(i,n) for(int i=0;i<n;++i)
#define F(i,j,n) for(int i=j;i<=n;++i)
#define D(i,j,n) for(int i=j;i>=n;--i)
#define pb push_back
#define CC(a,b) memset(a,b,sizeof(a))
using namespace std;
int getint(){
    int v=0,sign=1; char ch=getchar();
    while(!isdigit(ch)) {if(ch=='-') sign=-1; ch=getchar();}
    while(isdigit(ch))  {v=v*10+ch-'0'; ch=getchar();}
    return v*sign;
}
const int N=10000,M=100000,INF=~0u>>2;
typedef long long LL;
const double eps=1e-8;
/*******************template********************/
int n,m,w[N],l[N],c[N],d[N],du[N];
LL ans;
struct edge{int from,to,v,c;};
struct Net{
    edge E[M];
    int head[N],next[M],cnt;
    void ins(int x,int y,int z,int c){
        E[++cnt]=(edge){x,y,z,c};
        next[cnt]=head[x]; head[x]=cnt;
    }
    void add(int x,int y,int z,int c){
        ins(x,y,z,c); ins(y,x,0,-c);
    }
    int S,T,d[N],from[N],Q[M];
    bool inq[N];
    bool spfa(){
        int l=0,r=-1;
        F(i,0,T)d[i]=INF;
        d[S]=0; Q[++r]=S; inq[S]=1;
        while(l<=r){
            int x=Q[l++]; inq[x]=0;
            for(int i=head[x];i;i=next[i])
                if(E[i].v && d[x]+E[i].c<d[E[i].to]){
                    d[E[i].to]=d[x]+E[i].c;
                    from[E[i].to]=i;
                    if(!inq[E[i].to]){
                        Q[++r]=E[i].to;
                        inq[E[i].to]=1;
                    }
                }
        }
        return d[T]!=INF;
    }
    void mcf(){
        int x=INF;
        for(int i=from[T];i;i=from[E[i].from])
            x=min(x,E[i].v);
        for(int i=from[T];i;i=from[E[i].from]){
            E[i].v-=x;
            E[i^1].v+=x;
        }
        ans+=x*d[T];
    }
    void init(){
        n=getint(); m=getint(); cnt=1; ans=0;
        S=0; T=n+m+1;
        F(i,1,n){
            w[i]=getint();l[i]=getint();
            c[i]=getint();d[i]=getint();
//          ans+=w[i]*w[i]*c[i]+l[i]*l[i]*d[i];
        }
        int x,y;
        F(i,1,m){
            x=getint(); y=getint();
            du[x]++; du[y]++;
            add(x,i+n,1,0); add(y,i+n,1,0);
            add(i+n,T,1,0);
            l[x]++; l[y]++;
        }
        F(i,1,n) ans+=w[i]*w[i]*c[i]+l[i]*l[i]*d[i];
         
        F(i,1,n) F(j,1,du[i])
            add(S,i,1,((j+w[i])*2-1)*c[i] - ((l[i]-j+1)*2-1)*d[i] );
        while(spfa()) mcf();
        printf("%lld
",ans);
    }
}G1;
int main(){
#ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
//  freopen("output.txt","w",stdout);
#endif
    G1.init();
    return 0;
}

1449: [JSOI2009]球队收益

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 516  Solved: 283
[Submit][Status][Discuss]

Description

Input

Output

一个整数表示联盟里所有球队收益之和的最小值。

Sample Input

3 3
1 0 2 1
1 1 10 1
0 1 3 3
1 2
2 3
3 1

Sample Output

43

HINT

Source

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