hdu4295 4 substrings problem 状压dp

4 substrings problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 583    Accepted Submission(s): 180

Problem Description

　　One day you heard the following joke 
　　　　In America, you write strings.
　　　　In Soviet Russia, String writes YOU!! ¹
　　And find that now string is writting you! So to get rid of it, you must solve the following problem:
　　Given a string S and its four substring a,b,c, and d. In a configuration, you can place four substrings exactly one position it occurs in S (they may overlap), and characters covered by at least one such substring is called “covered”.
　　You should solve for minimum and maximum possible number of covered characters in a configuration.
　　You may assume that s contains only lowercase letters, and is of length less than 4096. However, lengths of a,b,c, and d would never exceed 64.
-------------------------------------------------------------------
¹An infamous Russian reversal

 

Input

　　There are several test cases.
　　For each test case there are 5 lines, denoting S,a,b,c, and d, respectively.
　　Please process until the EOF (End Of File).

 

Output

　　For each test case, please print a single line with two integers, first the minimum, then the maximum.

 

Sample Input

hellohelloabacabaabbaac

 

Sample Output

4 54 6

 

Source

2012 ACM/ICPC Asia Regional Chengdu Online

 

Recommend

liuyiding

 

状压大法好

先暴力处理每个字串在原串中的位置。。

用dp[i][j][k]表示主串选到第i位了，a，b，c，d四个字串有没有选过(2^4),从这一位开始后面已经放定的长度为k，表示的最大值或最小值

然后用队列维护一下即可。。。

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#define max(a,b) (a>=b ? a:b)#define min(a,b) (a<=b ? a:b)#define inf 0x3f3f3f3fusing namespace std;char s[4100],ss[4][66];bool check[4][4100];int len[4];void read(){    for(int i=0;i<4;i++) scanf("%s",ss[i]+1);}int dp[4100][16][66];struct node{    int x,st,l;    node(int x,int st,int l):x(x),st(st),l(l){}};int bfs(){    queue<node> q;    q.push(node(0,0,0));    memset(dp,-1,sizeof dp);    dp[0][0][0]=0;    int l=strlen(s+1);    while(!q.empty()){        node v=q.front();        q.pop();        if(v.x==l+1) continue;        for(int i=0;i<4;i++){            if(v.st&(1<<i)) continue;            int ll;            if(check[i][v.x]){                ll=max(v.l,len[i]);                if(dp[v.x][v.st|(1<<i)][ll]==-1){                    q.push(node(v.x,v.st|(1<<i),ll));                }                dp[v.x][v.st|(1<<i)][ll]=max(dp[v.x][v.st|(1<<i)][ll],dp[v.x][v.st][v.l]);            }        }        int ll=max(v.l-1,0);        if(dp[v.x+1][v.st][ll]==-1){            q.push(node(v.x+1,v.st,ll));        }        dp[v.x+1][v.st][ll]=max(dp[v.x+1][v.st][ll],dp[v.x][v.st][v.l]+(v.l ? 1:0));    }    int ret=-1;    for(int i=1;i<=l;i++){        ret=max(ret,dp[i][15][1]+1);    }    if(ret==-1) return 0;    else return ret;}int bfs1(){    queue<node> q;    q.push(node(0,0,0));    memset(dp,0x3f,sizeof dp);    dp[0][0][0]=0;    int l=strlen(s+1);    while(!q.empty()){        node v=q.front();        q.pop();        //cout<<v.x<<" "<<v.st<<" "<<v.l<<" "<<dp[v.x][v.st][v.l]<<endl;        if(v.x==l+1) continue;        for(int i=0;i<4;i++){            if(v.st&(1<<i)) continue;            int ll;            if(check[i][v.x]){                ll=max(v.l,len[i]);                if(dp[v.x][v.st|(1<<i)][ll]==inf){                    q.push(node(v.x,v.st|(1<<i),ll));                }                dp[v.x][v.st|(1<<i)][ll]=min(dp[v.x][v.st|(1<<i)][ll],dp[v.x][v.st][v.l]);            }        }        int ll=max(v.l-1,0);        if(dp[v.x+1][v.st][ll]==inf){            q.push(node(v.x+1,v.st,ll));        }        dp[v.x+1][v.st][ll]=min(dp[v.x+1][v.st][ll],dp[v.x][v.st][v.l]+(v.l ? 1:0));    }    int ret=inf;    for(int i=1;i<=l;i++){        ret=min(ret,dp[i][15][1]+1);    }    if(ret==inf) return 0;    else return ret;}void solve(){    memset(check,0,sizeof check);    int l=strlen(s+1);    for(int i=0;i<4;i++){        len[i]=strlen(ss[i]+1);        for(int j=1;j<=l;j++){            if(s[j]==ss[i][1]){                int k=1,tmp=j,jj=j;                while(k<=len[i]&&s[jj]==ss[i][k]) jj++,k++;                if(k==len[i]+1) {                    check[i][tmp]=1;                }            }        }    }    printf("%d %d\n",bfs1(),bfs());}int main(){    //freopen("input","r",stdin);    //freopen("out2","w",stdout);    while(~scanf("%s",s+1)){        read();        solve();    }    return 0;}