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  • 【Luogu】P2604网络扩容(费用流乱搞)

      题目链接

      这题比较水,就是乱改改费用流模板。判断一下已经满流的边和没有满流的边,然后再改改最大流模板,然后把它们拼起来就是了。

      话说这题第一遍90,然后撕烤一会发现自己yy的spfa扩容方式不允许反悔。然后改了一个貌似没什么用的地方,结果A了……

      但是改的那个地方貌似是真的没什么用啊……

    #include<cstdio>
    #include<cstdlib>
    #include<cctype>
    #include<algorithm>
    #include<cstring>
    #include<queue>
    #define maxn 2000
    #define maxm 10000
    using namespace std;
    
    inline long long read(){
        long long num=0,f=1;
        char ch=getchar();
        while(!isdigit(ch)){
            if(ch=='-')    f=-1;
            ch=getchar();
        }
        while(isdigit(ch)){
            num=num*10+ch-'0';
            ch=getchar();
        }
        return num*f;
    }
    
    inline int count(int i){    return i&1?i+1:i-1;    }
    
    struct Edge{
        int from,next,to,dis,val,flow;
    }edge[maxm*2];
    int head[maxn],num;
    inline void addedge(int from,int to,int dis,int val){
        edge[++num]=(Edge){from,head[from],to,dis,val,0};
        head[from]=num;
    }
    inline void add(int from,int to,int dis,int val){
        addedge(from,to,dis,val);
        addedge(to,from,-dis,0);
    }
    
    int dfn[maxn];
    bool vis[maxn];
    int Start,End;
    int list[maxn];
    
    bool bfs(){
        memset(vis,0,sizeof(vis));
        dfn[Start]=1;    vis[Start]=1;
        queue<int>q;    q.push(Start);
        while(!q.empty()){
            int from=q.front();q.pop();
            for(int i=head[from];i;i=edge[i].next){
                int to=edge[i].to;
                if(edge[i].val==edge[i].flow||vis[to])    continue;
                vis[to]=1;
                dfn[to]=dfn[from]+1;    q.push(to);
            }
        }
        return vis[End];
    }
    
    int dfs(int x,int val){
        if(x==End||val==0)    return val;
        int flow=0;    vis[x]=1;
        for(int &i=list[x];i;i=edge[i].next){
            int to=edge[i].to;
            if(edge[i].val==edge[i].flow||vis[to]||dfn[to]!=dfn[x]+1)    continue;
            int now=dfs(to,min(val,edge[i].val-edge[i].flow));
            edge[i].flow+=now;    edge[count(i)].flow-=now;    val-=now;    flow+=now;
            if(val<=0)    break;
        }
        if(flow!=val)    dfn[x]=-1;
        return flow;
    }
    
    int dis[maxn];
    int pre[maxn];
    
    int spfa(){
        memset(dis,127/3,sizeof(dis));    dis[Start]=0;
        memset(vis,0,sizeof(vis));
        queue<int>q;    q.push(Start);
        while(!q.empty()){
            
            int from=q.front();    q.pop();    vis[from]=0;
            for(int i=head[from];i;i=edge[i].next){
                int to=edge[i].to;
                if(edge[i].val>edge[i].flow){
                    if(dis[to]<=dis[from])    continue;
                    dis[to]=dis[from];
                }
                else{
                    if(edge[i].dis<0)    continue;
                    if(dis[to]<=dis[from]+edge[i].dis)    continue;
                    dis[to]=dis[from]+edge[i].dis;
                }
                pre[to]=i;
                if(vis[to])    continue;
                vis[to]=1;    q.push(to);
            }
        }
        int now=End;
        while(now!=Start&&now){
            int ret=pre[now];
            if(edge[ret].val==edge[ret].flow)    edge[ret].val++;
            edge[ret].flow++;    edge[count(ret)].flow--;
            now=edge[ret].from;
        }
        return dis[End];
    }
    
    int main(){
        int n=read(),m=read(),e=read();End=n;Start=1;
        for(int i=1;i<=m;++i){
            int a=read(),b=read(),c=read(),d=read();
            add(a,b,d,c);
        }
        int ans=0;
        while(bfs()){
            memset(vis,0,sizeof(vis));
            for(int i=1;i<=n;++i)    list[i]=head[i];
            int now=dfs(Start,0x7fffffff);
            if(!now)    break;
            ans+=now;
        }
        printf("%d",ans);
        long long now=0;
        for(int i=1;i<=e;++i)
            now+=spfa();
        printf(" %lld",now);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/cellular-automaton/p/8250665.html
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