最小费用可行流

https://www.cnblogs.com/guapisolo/p/10348428.html  原博

// ccf 201812-5 管道清洁: 最小费用可行流

#include<iostream>
#include<queue>
#include<string.h>
using namespace std;
const int maxn = 233;
const int inf = 0x3f3f3f3f;
struct node{
    int u,v,f,w,nxt;
    node(){};
    node(int a,int b,int c,int d,int e):u(a),v(b),f(c),w(d),nxt(e){}
}e[maxn*maxn];
int cnt = 0;
int head[maxn];
void add(int u,int v,int f,int w){
    e[cnt] = node(u,v,f,w,head[u]);
    head[u] = cnt++;
    e[cnt] = node(v,u,0,-w,head[v]);
    head[v] = cnt++;
}
int st,ed,E,ex;
int n,m;
int du[maxn];
int sum;
int num = 0;
void init(){
    scanf("%d%d",&n,&m);
    st = 0;ed = n + 1;
    ex = n + 2;
    cnt = 0;
    sum = 0;
    num = 0;
    memset(head,-1,ex<<2);
    memset(du,0,ex<<2);
    while(m--){
        int u,v;char c[2];
        scanf("%d%d%s",&u,&v,c);
        if(c[0] == 'A'){
            add(u,v,inf,E);
            du[u]--;
            du[v]++;
            num+=E;
        }
        else if(c[0] == 'B'){
            du[u]--;
            du[v]++;
            num+=E;
        }
        else if(c[0] == 'C'){
            add(u,v,inf,E);
        }
        else add(u,v,1,E);
    }
    for(int i = 1;i <= n;++i){
        if(du[i] > 0){
            sum += du[i];
            add(st,i,du[i],0);
        }
        else if(du[i] < 0){
            add(i,ed,-du[i],0);
        }
    }
}
int pre[maxn];
int flow[maxn];
int inq[maxn];
int dis[maxn];
int spfa(){
    memset(pre,-1,ex<<2);
    memset(inq,0,ex<<2);
    memset(dis,0x3f,ex<<2);
    queue<int> q;q.push(st);
    flow[st] = inf;inq[st] = 1;
    dis[st] = 0;
    while(q.size()){
        int u = q.front();q.pop();
        inq[u] = 0;
        for(int i = head[u];i != -1; i = e[i].nxt){
            int v = e[i].v;
            if(e[i].f && dis[v] > dis[u] + e[i].w){
                dis[v] = dis[u] + e[i].w;
                pre[v] = i;
                flow[v] = min(flow[u],e[i].f);
                if(!inq[v]) inq[v] = 1,q.push(v);
            }
        }
    }
    if(pre[ed] == -1) return -1;
    return flow[ed];
}
int mfmv(){
    int fw = 0;
    int ans = 0;
    int d;
    while((d = spfa())!=-1){
        fw += d;
        ans += dis[ed]*d;
        int v = ed;
        while(v!=st){
            e[pre[v]].f -= d;
            e[pre[v]^1].f += d;
            v = e[pre[v]].u;
        }
    }
    if(fw!=sum) return -1;
    return ans + num;
}
void sol(){
    int ans = mfmv();
    printf("%d\n",ans);
}
int main(){
    int T;cin>>T;cin>>E>>E;
    while(T--){
        init();sol();
    }
}