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=0;
    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]=0;
    queue<int>que;
    que.push(vs);
    while(!que.empty()){
        int u=que.front();
        que.pop();
        mark[u]=false;
        for(int i=head[u];i!=-1;i=edge[i].next){
            int v=edge[i].v,cost=edge[i].cost;
            if(edge[i].cap>0&&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=0,cost=0;
    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]^1].cap+=aug;
        }
    }
    printf("%d %d
",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,0,sizeof(g));
        for(int i=0;i<len-1;i++){
            int u=str1[i]-'a',v=str1[i+1]-'a';
            if(!g[u][v]){
                g[u][v]=(i+1)*(i+1);
            }
        }
        scanf("%s",str2);
        len=strlen(str2);
        vs=0,vt=2*len+1;
        NE=0;
        memset(head,-1,sizeof(head));
        for(int i=0;i<len;i++){
            Insert(vs,i+1,1,0);
            Insert(i+1+len,vt,1,0);
            for(int j=i+1;j<len;j++){
                int u=str2[i]-'a',v=str2[j]-'a';
                if(g[u][v]){
                    Insert(i+1,j+1+len,1,g[u][v]);
                }
            }
        }
        MinCostFlow(vs,vt);
    }
    return 0;
}