uva 11381(神奇的构图、最小费用最大流）

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=29821

思路：首先拆点，每个字符对应的位置拆成i和i+len，然后源点和i连边，容量为1,花费为0，i+len与汇点连边,容量为1,花费为0，然后就是对于那些在P中的字符串连边，容量为1，花费为g[u][v],最后跑最小费最大流即可，然后答案就是len-flow,cost;

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<climits>
 #include<cstdlib>
 #include<algorithm>
 #include<stack>
 #include<vector>
 #include<queue>
 #include<string>
 #include<cmath>
 #include<set>
 using namespace std;
 #define inf 1<<30
 #define INF 1LL<<60
 typedef long long ll;
 typedef pair<int,int>PP;
 #define MAXN 1111

 struct Edge{
     int v,cap,cost,next;
 }edge[MAXN*MAXN];

 int len,vs,vt,NE;
 int head[MAXN];

 void Insert(int u,int v,int cap,int cost)
 {
     edge[NE].v=v;
     edge[NE].cap=cap;
     edge[NE].cost=cost;
     edge[NE].next=head[u];
     head[u]=NE++;

     edge[NE].v=u;
     edge[NE].cap=;
     edge[NE].cost=-cost;
     edge[NE].next=head[v];
     head[v]=NE++;
 }

 int pre[MAXN],cur[MAXN];
 bool mark[MAXN];
 int dist[MAXN];

 bool spfa(int vs,int vt)
 {
     memset(mark,false,sizeof(mark));
     fill(dist,dist+MAXN,inf);
     dist[vs]=;
     queue<int>que;
     que.push(vs);
     while(!que.empty()){
         int u=que.front();
         que.pop();
         mark[u]=false;
         for(int i=head[u];i!=-;i=edge[i].next){
             int v=edge[i].v,cost=edge[i].cost;
             if(edge[i].cap>&&dist[u]+cost<dist[v]){
                 dist[v]=dist[u]+cost;
                 pre[v]=u;
                 cur[v]=i;
                 if(!mark[v]){
                     mark[v]=true;
                     que.push(v);
                 }
             }
         }
     }
     return dist[vt]<inf;
 }

 void MinCostFlow(int vs,int vt)
 {
     int flow=,cost=;
     while(spfa(vs,vt)){
         int aug=inf;
         for(int u=vt;u!=vs;u=pre[u]){
             aug=min(aug,edge[cur[u]].cap);
         }
         flow+=aug,cost+=dist[vt]*aug;
         for(int u=vt;u!=vs;u=pre[u]){
             edge[cur[u]].cap-=aug;
             edge[cur[u]^].cap+=aug;
         }
     }
     printf("%d %d\n",len-flow,cost);
 }

 char str1[MAXN],str2[MAXN];
 int g[MAXN][MAXN];
 int main()
 {
     int _case;
     scanf("%d",&_case);
     while(_case--){
         scanf("%s",str1);
         len=strlen(str1);
         memset(g,,sizeof(g));
         for(int i=;i<len-;i++){
             int u=str1[i]-'a',v=str1[i+]-'a';
             if(!g[u][v]){
                 g[u][v]=(i+)*(i+);
             }
         }
         scanf("%s",str2);
         len=strlen(str2);
         vs=,vt=*len+;
         NE=;
         memset(head,-,sizeof(head));
         for(int i=;i<len;i++){
             Insert(vs,i+,,);
             Insert(i++len,vt,,);
             for(int j=i+;j<len;j++){
                 int u=str2[i]-'a',v=str2[j]-'a';
                 if(g[u][v]){
                     Insert(i+,j++len,,g[u][v]);
                 }
             }
         }
         MinCostFlow(vs,vt);
     }
     return ;
 }