【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=,sign=; char ch=getchar();

     while(!isdigit(ch)) {if(ch=='-') sign=-; ch=getchar();}

     while(isdigit(ch))  {v=v*+ch-''; ch=getchar();}

     return v*sign;

 }

 const int N=,M=,INF=~0u>>;

 typedef long long LL;

 const double eps=1e-;

 /*******************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,,-c);

     }

     int S,T,d[N],from[N],Q[M];

     bool inq[N];

     bool spfa(){

         int l=,r=-;

         F(i,,T)d[i]=INF;

         d[S]=; Q[++r]=S; inq[S]=;

         while(l<=r){

             int x=Q[l++]; inq[x]=;

             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]=;

                     }

                 }

         }

         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^].v+=x;

         }

         ans+=x*d[T];

     }

     void init(){

         n=getint(); m=getint(); cnt=; ans=;

         S=; T=n+m+;

         F(i,,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,,m){

             x=getint(); y=getint();

             du[x]++; du[y]++;

             add(x,i+n,,); add(y,i+n,,);

             add(i+n,T,,);

             l[x]++; l[y]++;

         }

         F(i,,n) ans+=w[i]*w[i]*c[i]+l[i]*l[i]*d[i];

         F(i,,n) F(j,,du[i])

             add(S,i,,((j+w[i])*-)*c[i] - ((l[i]-j+)*-)*d[i] );

         while(spfa()) mcf();

         printf("%lld\n",ans);

     }

 }G1;

 int main(){

 #ifndef ONLINE_JUDGE

     freopen("input.txt","r",stdin);

 //  freopen("output.txt","w",stdout);

 #endif

     G1.init();

     return ;

 }

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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