题解：
戳这里：http://blog.csdn.net/huzecong/article/details/9119741?
代码：
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set> 10 #include<queue> 11 #include<string> 12 #define inf 1000000000 13 #define maxn 100000+5 14 #define maxm 100000+5 15 #define eps 1e-10 16 #define ll long long 17 #define pa pair<int,int> 18 #define for0(i,n) for(int i=0;i<=(n);i++) 19 #define for1(i,n) for(int i=1;i<=(n);i++) 20 #define for2(i,x,y) for(int i=(x);i<=(y);i++) 21 #define for3(i,x,y) for(int i=(x);i>=(y);i--) 22 #define for4(i,x) for(int i=head[x],y;i;i=e[i].next) 23 #define mod 1000000007 24 using namespace std; 25 inline int read() 26 { 27 int x=0,f=1;char ch=getchar(); 28 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 29 while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();} 30 return x*f; 31 } 32 int n,m,k,mincost,tot=1,s,t,a[maxn],b[maxn],c[maxn],d[maxn],ss[maxn],head[maxn],from[2*maxm]; 33 bool v[maxn]; 34 queue<int>q; 35 struct edge{int from,go,next,v,c;}e[2*maxm]; 36 void ins(int x,int y,int z,int w) 37 { 38 e[++tot]=(edge){x,y,head[x],z,w};head[x]=tot; 39 } 40 void insert(int x,int y,int z,int w) 41 { 42 ins(x,y,z,w);ins(y,x,0,-w); 43 } 44 bool spfa() 45 { 46 for (int i=s;i<=t;i++){v[i]=0;d[i]=inf;} 47 q.push(s);d[s]=0;v[s]=1; 48 while(!q.empty()) 49 { 50 int x=q.front();q.pop();v[x]=0; 51 for (int i=head[x],y;i;i=e[i].next) 52 if(e[i].v&&d[x]+e[i].c<d[y=e[i].go]) 53 { 54 d[y]=d[x]+e[i].c;from[y]=i; 55 if(!v[y]){v[y]=1;q.push(y);} 56 } 57 } 58 return d[t]!=inf; 59 } 60 void mcf() 61 { 62 while(spfa()) 63 { 64 int tmp=inf; 65 for(int i=from[t];i;i=from[e[i].from]) tmp=min(tmp,e[i].v); 66 mincost+=d[t]*tmp; 67 for(int i=from[t];i;i=from[e[i].from]){e[i].v-=tmp;e[i^1].v+=tmp;} 68 } 69 } 70 int main() 71 { 72 freopen("input.txt","r",stdin); 73 freopen("output.txt","w",stdout); 74 n=read();m=read();s=0;t=n+m+1; 75 for1(i,n)a[i]=read(),b[i]=read(),c[i]=read(),d[i]=read(); 76 for1(i,m) 77 { 78 int x=read(),y=read();ss[x]++;ss[y]++; 79 insert(s,i+n,1,0); 80 insert(i+n,x,1,0); 81 insert(i+n,y,1,0); 82 } 83 for1(i,n)b[i]+=ss[i]; 84 for1(i,n) 85 { 86 int x=a[i],y=b[i]; 87 for1(j,ss[i]) 88 { 89 insert(i,t,1,2*c[i]*x-2*d[i]*y+c[i]+d[i]); 90 x++;y--; 91 } 92 } 93 int ans=0; 94 for1(i,n)ans+=c[i]*a[i]*a[i]+d[i]*b[i]*b[i]; 95 mcf(); 96 printf("%d ",ans+mincost); 97 return 0; 98 }
1449: [JSOI2009]球队收益
Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 411 Solved: 225
[Submit][Status]
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
1 0 2 1
1 1 10 1
0 1 3 3
1 2
2 3
3 1
Sample Output
43
HINT
