【传送门:BZOJ3511】
简要题意:
给出n个点,m条边,每个点有A和B两种形态,一开始1为A,n为B
给出VA[i]和VB[i],表示第i个点选择A和B形态的价值
每条边给出x,y,EA,EB,EC,表示如果x和y都为A,则获得EA价值,如果都为B则获得EB价值,否则会得到EC的费用(就是负价值)
求出最大价值
题解:
神奇的最小割,太强了
建图膜
参考代码:
#include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<cmath> using namespace std; struct node {int x,y,c,next,other; }a[2100000];int len,last[110000]; void ins(int x,int y,int c) {int k1=++len,k2=++len;a[k1].x=x;a[k1].y=y;a[k1].c=c;a[k1].next=last[x];last[x]=k1;a[k2].x=y;a[k2].y=x;a[k2].c=0;a[k2].next=last[y];last[y]=k2;a[k1].other=k2;a[k2].other=k1; } int h[110000],list[110000],st,ed; bool bt_h() {memset(h,0,sizeof(h));h[st]=1;int head=1,tail=2;list[1]=st;while(head!=tail){int x=list[head];for(int k=last[x];k;k=a[k].next){int y=a[k].y;if(h[y]==0&&a[k].c>0){h[y]=h[x]+1;list[tail++]=y;}}head++;}if(h[ed]==0) return false;else return true; } int findflow(int x,int f) {if(x==ed) return f;int s=0,t;for(int k=last[x];k;k=a[k].next){int y=a[k].y;if(h[y]==(h[x]+1)&&a[k].c>0&&f>s){t=findflow(y,min(a[k].c,f-s));s+=t;a[k].c-=t;a[a[k].other].c+=t;}}if(s==0) h[x]=0;return s; } int main() {int n,m;scanf("%d%d",&n,&m);int sum=0;st=0;ed=n+2*m+1;len=0;memset(last,0,sizeof(last));ins(st,1,999999999);for(int i=2;i<n;i++){int d;scanf("%d",&d);sum+=d;ins(st,i,d);}ins(n,ed,999999999);for(int i=2;i<n;i++){int d;scanf("%d",&d);sum+=d;ins(i,ed,d);}for(int i=1;i<=m;i++){int x,y,ea,eb,ec;scanf("%d%d%d%d%d",&x,&y,&ea,&eb,&ec);sum+=ea+eb;ins(x,y,ec);ins(y,x,ec);ins(i+n,x,999999999);ins(i+n,y,999999999);ins(x,i+n+m,999999999);ins(y,i+n+m,999999999);ins(st,i+n,ea);ins(i+n+m,ed,eb);}while(bt_h()==true) sum-=findflow(st,999999999);printf("%d\n",sum);return 0; }