//最小费用最大流 //源点向每种字母连边,容量为t中该种字母的需求 //每种字母向每个字符串连边,容量为每个字符串自身有多少该种字母 //每个字符串向汇点连边,容量为无限大,费用为字符串自身的长度 //看最大流量是不是等于t的长度,如果等于则输出Yes,最少费用 #include<bits/stdc++.h> using namespace std; const int maxn=2e5+100; const int inf=1e9; int n; string S[maxn]; int cnt[256][maxn]; int s,t; int visit[maxn]; int dis[maxn];//源点到每个点的最小花费(最短路) int pre[maxn];//前驱 int lst[maxn];//每个点所连的前一条边 int flow[maxn];//源点到此处的流量 int maxflow;//最大流 int mincost;//最小费用 struct node { int u,v,flow,dis,nxt; }edge[maxn*2]; int tot; int head[maxn]; void addedge (int u,int v,int flow,int dis) { edge[tot].u=u; edge[tot].v=v; edge[tot].flow=flow; edge[tot].dis=dis; edge[tot].nxt=head[u]; head[u]=tot++; edge[tot].u=v; edge[tot].v=u; edge[tot].flow=0; edge[tot].dis=-dis; edge[tot].nxt=head[v]; head[v]=tot++; } bool spfa (int s,int t) { for (int i=0;i<maxn;i++) dis[i]=inf,flow[i]=inf,visit[i]=0; queue<int> q; q.push(s); visit[s]=1; dis[s]=0; pre[t]=-1; while (!q.empty()) { int u=q.front(); q.pop(); visit[u]=0; for (int i=head[u];i!=-1;i=edge[i].nxt) { if (edge[i].flow>0&&dis[edge[i].v]>dis[u]+edge[i].dis) { dis[edge[i].v]=dis[u]+edge[i].dis; pre[edge[i].v]=u; lst[edge[i].v]=i; flow[edge[i].v]=min(flow[u],edge[i].flow); if (!visit[edge[i].v]) { visit[edge[i].v]=1; q.push(edge[i].v); } } } } return pre[t]!=-1; } void MCMF () { while (spfa(s,t)) { int u=t; maxflow+=flow[t]; mincost+=dis[t]; while (u!=s) { edge[lst[u]].flow-=flow[t]; edge[lst[u]^1].flow+=flow[t]; u=pre[u]; } } } int main () { cin>>S[0]; scanf("%d",&n); for (int i=0;i<maxn;i++) head[i]=-1; for (int i=1;i<=n;i++) cin>>S[i]; for (int i=0;i<=n;i++) { for (int j=0;j<S[i].length();j++) { cnt[S[i][j]][i]++; } } //源点是0 //1~256是每种字母 //256+1~256+1+n是每种字符串 //256+1+n+1是汇点 s=0; t=258+n; for (int i=1;i<=256;i++) addedge(s,i,cnt[i-1][0],0); for (int i=1;i<=256;i++) { for (int j=1;j<=n;j++) { addedge(i,j+256,cnt[i-1][j],0); } } for (int i=1;i<=n;i++) addedge(256+i,t,inf,S[i].length()); MCMF(); if (maxflow<S[0].length()) { printf("No %d ",S[0].length()-maxflow); } else { int ans=0; for (int i=0;i<tot;i+=2) if (edge[i^1].flow) ans+=edge[i].dis; printf("Yes %d ",ans-S[0].length()); } }