Educational Codeforces Round 55 G 最小割

题意,一个图,选一个子图,使得 max(\sum v[i] -\sum a[i]) v[i] 边权,a[i]点权

思路:

最小割经典模型,我们把边和源相连流量v[i],点和汇向连流量a[i],中间的依赖用inf相连,那么这样的图最小割的含义就是“不选的边和选了的点的和”,那么这个值最小答案自然最大,\sum v[i] - maxflow 就是答案了

代码:

#include
#define PB push_back
#define X first
#define Y second
#define FIO std::ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
typedef long long ll;
typedef double LD;
typedef pair pii;
const int maxn=5e5+10;
const ll inf=1e10+7;
const int MAXN=3100;
ll maze[MAXN][MAXN];
ll gap[MAXN],dis[MAXN],pre[MAXN],cur[MAXN];
ll sap(int start,int end,int nodenum){
    memset(cur,0,sizeof(cur));
    memset(dis,0,sizeof(dis));
    memset(gap,0,sizeof(gap));
    ll u=pre[start]=start,maxflow=0,aug=-1;
    gap[0]=nodenum;
    while(dis[start]maze[u][v])aug=maze[u][v];
                pre[v]=u;
                u=cur[u]=v;
                if(v==end){
                    maxflow+=aug;
                    for(u=pre[u]; v!=start; v=u,u=pre[u]){
                        maze[u][v]-=aug;
                        maze[v][u]+=aug;
                    }
                    aug=-1;
                }
                goto loop;
            }
        int mindis=nodenum-1;
        for(int v=0; vdis[v]){
                cur[u]=v;
                mindis=dis[v];
            }
        if((--gap[dis[u]])==0)break;
        gap[dis[u]=mindis+1]++;
        u=pre[u];
    }
    return maxflow;
}
int n,m,x,y,z;
ll a[MAXN];
int main(){
    FIO;
    cin>>n>>m;
    ll ans=0;
    for(int i=1;i<=n;i++)cin>>a[i],maze[i][n+m+2]=a[i];
    for(int i=1;i<=m;i++){
        cin>>x>>y>>z;
        ans+=z;
        maze[n+i][x]=inf;
        maze[n+i][y]=inf;
        maze[0][n+i]=z;
    }
    cout<

 

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