【BZOJ】【1449】【JSOI2009】球队收益

网络流/费用流/二分图最小权匹配


  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

  1 /**************************************************************
  2     Problem: 1449
  3     User: Tunix
  4     Language: C++
  5     Result: Accepted
  6     Time:676 ms
  7     Memory:3940 kb
  8 ****************************************************************/
  9  
 10 //BZOJ 1449
 11 #include<cmath>
 12 #include<vector>
 13 #include<cstdio>
 14 #include<cstring>
 15 #include<cstdlib>
 16 #include<iostream>
 17 #include<algorithm>
 18 #define rep(i,n) for(int i=0;i<n;++i)
 19 #define F(i,j,n) for(int i=j;i<=n;++i)
 20 #define D(i,j,n) for(int i=j;i>=n;--i)
 21 #define pb push_back
 22 #define CC(a,b) memset(a,b,sizeof(a))
 23 using namespace std;
 24 int getint(){
 25     int v=0,sign=1; char ch=getchar();
 26     while(!isdigit(ch)) {if(ch=='-') sign=-1; ch=getchar();}
 27     while(isdigit(ch))  {v=v*10+ch-'0'; ch=getchar();}
 28     return v*sign;
 29 }
 30 const int N=10000,M=100000,INF=~0u>>2;
 31 typedef long long LL;
 32 const double eps=1e-8;
 33 /*******************template********************/
 34 int n,m,w[N],l[N],c[N],d[N],du[N];
 35 LL ans;
 36 struct edge{int from,to,v,c;};
 37 struct Net{
 38     edge E[M];
 39     int head[N],next[M],cnt;
 40     void ins(int x,int y,int z,int c){
 41         E[++cnt]=(edge){x,y,z,c};
 42         next[cnt]=head[x]; head[x]=cnt;
 43     }
 44     void add(int x,int y,int z,int c){
 45         ins(x,y,z,c); ins(y,x,0,-c);
 46     }
 47     int S,T,d[N],from[N],Q[M];
 48     bool inq[N];
 49     bool spfa(){
 50         int l=0,r=-1;
 51         F(i,0,T)d[i]=INF;
 52         d[S]=0; Q[++r]=S; inq[S]=1;
 53         while(l<=r){
 54             int x=Q[l++]; inq[x]=0;
 55             for(int i=head[x];i;i=next[i])
 56                 if(E[i].v && d[x]+E[i].c<d[E[i].to]){
 57                     d[E[i].to]=d[x]+E[i].c;
 58                     from[E[i].to]=i;
 59                     if(!inq[E[i].to]){
 60                         Q[++r]=E[i].to;
 61                         inq[E[i].to]=1;
 62                     }
 63                 }
 64         }
 65         return d[T]!=INF;
 66     }
 67     void mcf(){
 68         int x=INF;
 69         for(int i=from[T];i;i=from[E[i].from])
 70             x=min(x,E[i].v);
 71         for(int i=from[T];i;i=from[E[i].from]){
 72             E[i].v-=x;
 73             E[i^1].v+=x;
 74         }
 75         ans+=x*d[T];
 76     }
 77     void init(){
 78         n=getint(); m=getint(); cnt=1; ans=0;
 79         S=0; T=n+m+1;
 80         F(i,1,n){
 81             w[i]=getint();l[i]=getint();
 82             c[i]=getint();d[i]=getint();
 83 //          ans+=w[i]*w[i]*c[i]+l[i]*l[i]*d[i];
 84         }
 85         int x,y;
 86         F(i,1,m){
 87             x=getint(); y=getint();
 88             du[x]++; du[y]++;
 89             add(x,i+n,1,0); add(y,i+n,1,0);
 90             add(i+n,T,1,0);
 91             l[x]++; l[y]++;
 92         }
 93         F(i,1,n) ans+=w[i]*w[i]*c[i]+l[i]*l[i]*d[i];
 94          
 95         F(i,1,n) F(j,1,du[i])
 96             add(S,i,1,((j+w[i])*2-1)*c[i] - ((l[i]-j+1)*2-1)*d[i] );
 97         while(spfa()) mcf();
 98         printf("%lld\n",ans);
 99     }
100 }G1;
101 int main(){
102 #ifndef ONLINE_JUDGE
103     freopen("input.txt","r",stdin);
104 //  freopen("output.txt","w",stdout);
105 #endif
106     G1.init();
107     return 0;
108 }
View Code

 

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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posted @ 2015-03-20 11:30  Tunix  阅读(213)  评论(0编辑  收藏  举报