【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
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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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