hdu 5772 String problem 最大权闭合子图

String problem

题目连接:

http://acm.hdu.edu.cn/showproblem.php?pid=5772

Description

This is a simple problem about string. Now a string S contains only ‘0’-‘9’. ?? wants to select a subsequence from this string. And makes this subsequence score maximum. The subsequence’s score is calculated as follows:
Score= Value – Total_Cost
The calculation of the Cost is as follows:
If the number of characters x in the subsequence is kx, And the two coefficients are ax,bx,The cost of character x calculated as follows:

{cost[x]=0,kx=0cost[x]=ax∗(kx−1)+bx,kx≠0

TotalCost=∑i=09cost[i]

The calculation of the Value is as follows:

Value=0;
for(int i=1;i<=length(substr);++i){
for(int j=1;j<=length(substr);++j){
if(i!=j)
Value+=w[id[i]][id[j]];
}
}

id[i] is the position of the subsequence’s ith character in the original string,for example,if the original string is “13579”,and the subsubquence is “159”,then the array id ={1,3,5}. The w is a weight matrix.

Input

The first line contains an integer T, denoting the number of the test cases.
For each test case, the first line contains one integers n, the length of a string.
Next line contains the string S.
Next ten lines,each line contains ai,bi,denote the char i’s(0-9) coefficients
Next is a n*n matrix w.
Limits:
T<=20,
0<=n<=100
0<=ai<=bi<=1000
0<=w[i][j]<=50

Output

Each test output one line “Case #x: y” , where x is the case number ,staring from 1. y is the Maximum score.

Sample Input

1
3
135
1 2
1 2
1 2
1 2
1 2
1 2
1 2
1 2
1 2
1 2
0 0 3
1 0 0
4 0 0

Sample Output

Case #1: 3

Hint
we can choose “15”,id[]={1,3} then Value=w[1][3]+w[3][1]=7,
Total_Cost=2+2=4,Score=7-4=3

Hint

题意

给你一个字符串,只含有数字。

你需要选择出一个子序列,使得这个子序列的权值最大。

这个子序列如果这个数字第一次出现就ans-=bx,否则就-=ax

然后如果第i个字符和第j个字符都在子序列里面,那么ans+=w[i][j]

问你最大ans是多少

题解:

比较显然的就是最大闭合子图了,这个跑网络流就是了。

标准题解:

网络流:最大权闭合子图。
思路如下:

首先将点分为3类

第一类:Pij 表示第i个点和第j个点组合的点,那么Pij的权值等于w[i][j]+w[j][i](表示得分)

第二类:原串中的n个点每个点拆出一个点,第i个点权值为 –a[s[i]] (表示要花费)

第三类:对于10种字符拆出10个点,每个点的权值为 -(b[x]-a[x])

那么我们可以得到一个关系图 ,对于第一类中的点Pij,如果想要选择Pij,你就必须要选中第二类中的点i和j,对于第二类中的点如果你想选中第i个点,其对应的字符s[i],那么就必须选中第三类中s[i] 对应的点,因为每个种类的点第一次选中时花费是b[s[i]],而第二类中花费都是a[s[i]],一定要补上b[s[i]]-a[s[i]],而且只需要补上一次。

得到上面的关系图后然后就是普通的最大权闭合子图问题,直接求解即可。

但是我们队比较邪,我们先2^10枚举了一下,然后再跑的网络流,但是这样会TLE。

于是我们就XJB只搜了后面的几个状态和前面的状态,然后跑网络流就过了……

下面代码是我们队的谐星做法,非正解,正解按照前面的思路建图就好了。

另外这个图跑ISAP比跑dinic快的多。。。。

代码

#include<bits/stdc++.h>
#define Maxn 60009
#define Maxm 400009
#define inf 100000000
#define LL int
using namespace std;
const int MAXN=100000,MAXM=100000;
struct Edge
{
    int v,c,f,nx;
    Edge() {}
    Edge(int v,int c,int f,int nx):v(v),c(c),f(f),nx(nx) {}
} E[MAXM];
int G[MAXN],cur[MAXN],pre[MAXN],dis[MAXN],gap[MAXN],N,sz;
void init(int _n)
{
    N=_n,sz=0; memset(G,-1,sizeof(G[0])*N);
}
void link(int u,int v,int c)
{
    E[sz]=Edge(v,c,0,G[u]); G[u]=sz++;
    E[sz]=Edge(u,0,0,G[v]); G[v]=sz++;
}
int ISAP(int S,int T)
{//S -> T
    int maxflow=0,aug=inf,flag=false,u,v;
    for (int i=0;i<N;++i)cur[i]=G[i],gap[i]=dis[i]=0;
    for (gap[S]=N,u=pre[S]=S;dis[S]<N;flag=false)
    {
        for (int &it=cur[u];~it;it=E[it].nx)
        {
            if (E[it].c>E[it].f&&dis[u]==dis[v=E[it].v]+1)
            {
                if (aug>E[it].c-E[it].f) aug=E[it].c-E[it].f;
                pre[v]=u,u=v; flag=true;
                if (u==T)
                {
                    for (maxflow+=aug;u!=S;)
                    {
                        E[cur[u=pre[u]]].f+=aug;
                        E[cur[u]^1].f-=aug;
                    }
                    aug=inf;
                }
                break;
            }
        }
        if (flag) continue;
        int mx=N;
        for (int it=G[u];~it;it=E[it].nx)
        {
            if (E[it].c>E[it].f&&dis[E[it].v]<mx)
            {
                mx=dis[E[it].v]; cur[u]=it;
            }
        }
        if ((--gap[dis[u]])==0) break;
        ++gap[dis[u]=mx+1]; u=pre[u];
    }
    return maxflow;
}
bool bfs(int S,int T)
{
    static int Q[MAXN]; memset(dis,-1,sizeof(dis[0])*N);
    dis[S]=0; Q[0]=S;
    for (int h=0,t=1,u,v,it;h<t;++h)
    {
        for (u=Q[h],it=G[u];~it;it=E[it].nx)
        {
            if (dis[v=E[it].v]==-1&&E[it].c>E[it].f)
            {
                dis[v]=dis[u]+1; Q[t++]=v;
            }
        }
    }
    return dis[T]!=-1;
}
int dfs(int u,int T,int low)
{
    if (u==T) return low;
    int ret=0,tmp,v;
    for (int &it=cur[u];~it&&ret<low;it=E[it].nx)
    {
        if (dis[v=E[it].v]==dis[u]+1&&E[it].c>E[it].f)
        {
            if (tmp=dfs(v,T,min(low-ret,E[it].c-E[it].f)))
            {
                ret+=tmp; E[it].f+=tmp; E[it^1].f-=tmp;
            }
        }
    }
    if (!ret) dis[u]=-1; return ret;
}
int dinic(int S,int T)
{
    int maxflow=0,tmp;
    while (bfs(S,T))
    {
        memcpy(cur,G,sizeof(G[0])*N);
        while (tmp=dfs(S,T,inf)) maxflow+=tmp;
    }
    return maxflow;
}

struct st
{
    int u,v;
    LL value;
}e[Maxm];
int cost[MAXN];
int mp[105][105];
int a[10],b[10],nn;
int Ans = 0;
string s;
LL get(int zt){
    vector<int> tmp;
    LL sum=0;
    int cccc = 0;
    for(int i=0;i<10;i++){
        if((1<<i)&zt){
            sum-=(b[i]-a[i]);
        }
    }
    for(int i=0;i<s.size();i++){
        int num = s[i]-'0';
        if((1<<num)&zt){
            tmp.push_back(i+1);
            cost[i+1]=a[s[i]-'0'];
         //   cout<<i<<" ";
        }
    }
    int m = 1;

    for(int i=0;i<tmp.size();i++){
        for(int j=i+1;j<tmp.size();j++){
            e[m].u=tmp[i];
            e[m].v=tmp[j];
            e[m].value=mp[tmp[i]][tmp[j]]+mp[tmp[j]][tmp[i]];
            sum+=e[m].value;
            m++;
        }
    }
    if(sum<=Ans)return 0;
    m--;
    init(12000);
    int n = tmp.size();
    for(int i=1;i<=m;i++)
    {
        link(0,i,e[i].value);
        link(i,m+e[i].u,inf);
        link(i,m+e[i].v,inf);
    }
    for(int i=1;i<=n;i++)
        link(tmp[i-1]+m,m+nn+1,cost[tmp[i-1]]);
    LL ans=ISAP(0,m+nn+1);
    return sum-ans;
}
int cas = 0;
int vvv[15];
int ccc = 0;
int times = 0 ;

void dfs1(int x,int tmp,int k){
    if(x==10){

        if( (k>=ccc-7||ccc<=2) )Ans=max(Ans,get(tmp));
        return;
    }
    if(vvv[x])dfs1(x+1,tmp|(1<<x),k+1);
    dfs1(x+1,tmp,k);
}

void solve(){
    memset(vvv,0,sizeof(vvv));
    Ans=0;
    times=0;
    scanf("%d",&nn);
    cin>>s;
    ccc=0;
    for(int i=0;i<s.size();i++){
        if(vvv[s[i]-'0']==0)ccc++;
        vvv[s[i]-'0']=1;
    }
    for(int i=0;i<10;i++){
        scanf("%d%d",&a[i],&b[i]);
    }
    for(int i=1;i<=nn;i++){
        for(int j=1;j<=nn;j++){
            scanf("%d",&mp[i][j]);
        }
    }

    dfs1(0,0,0);


    cout<<"Case #"<<++cas<<": "<<Ans<<endl;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--)solve();
    return 0;
}
posted @ 2016-07-28 19:05 qscqesze 阅读(...) 评论(...) 编辑 收藏