HUNAN 11569 Just Another Knapsack Problem(AC自动机+dp)

http://acm.hunnu.edu.cn/online/?action=problem&type=show&id=11569&courseid=0

给出目标串,每个子串和对应的权值,然后要从子串中匹配出目标串并且权值最大.匹配的位置不能重复.

dp[i]为匹配到i这个位置时的最大价值,那么dp[i]=max(dp[i],dp[i-len[j]]+val[j]);

每次找到的匹配的子串在目标串的位置,然后动态转移。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))

#define ll long long
#define inf 0x7f7f7f7f
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)

#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define lowbit(x)   (x)&(-x)
#define Read()  freopen("a.txt", "r", stdin)
#define Write() freopen("dout.txt", "w", stdout);


#define Nn 510007
#define Mc 26

using namespace std;

class Acautomaton
{
    private:

    int chd[Nn][Mc];
    int fail[Nn];
    int ID[Mc];
    int val[Nn];
    int Q[Nn];
    int sz;
    int len[Nn];

    public:
    int dp[101000];
    void Init()
    {
        fail[0] = 0;
        for (int i = 0; i < Mc; ++i)
        {
            ID[i] = i;
        }
    }
    void Reset()
    {
        CL(chd[0],0);
        sz = 1;
    }
    int idx(char c) {return c-'a';}
    void insert(char *s,int key)
    {
        int p = 0;
        for (; *s; s++)
        {
            int k = ID[*s - 'a'];
            if (!chd[p][k])
            {
                CL(chd[sz],0);
                val[sz] = 0;
                len[sz]=0;
                chd[p][k] = sz++;
            }
            len[chd[p][k]]=len[p]+1;
            p = chd[p][k];
        }
        val[p] =max(val[p], key);
    }
    void Build()
    {
        int *s = Q ,*e = Q,i;
        for (i = 0; i < Mc; ++i)
        {
            if (chd[0][i])
            {
                *e++ = chd[0][i];
                fail[chd[0][i]] = 0;
            }
        }
        while (s != e)
        {
            int u = *s++;
            for (i = 0; i < Mc; ++i)
            {
                int &v = chd[u][i];
                if (v)
                {
                    *e++ = v;
                    fail[v] = chd[fail[u]][i];
                }
                else v = chd[fail[u]][i];
            }
        }
    }
    void find(char *T) {
        int j=0;
        for(int i=0,temp,c;T[i];i++) {
            dp[i]=0;
            c=idx(T[i]);
            while(j&&!chd[j][c]) j=fail[j];
            j=chd[j][c];
            temp=j;
            while(temp) {
                if(val[temp]) {
                    //printf("%d\n",val[temp]);
                    if(i-len[temp]<0) //temp已经在目标串中出现,但是不能转移 
                        dp[i]=max(dp[i],val[temp]);
                    else if(dp[i-len[temp]])
                        dp[i]=max(dp[i],dp[i-len[temp]]+val[temp]);
                    printf("%d %d %d\n",i,dp[i],len[temp]);
                }
                temp=fail[temp];
            }

        }
    }
   /* int solve(char *s)
    {
        int p = 0;
        int k,ans = 0;
        for (; *s; s++)
        {
            k = ID[*s - 'a'];
            while (!chd[p][k] && p != 0) p = fail[p];

            p = chd[p][k];

            int rt = p;
            while (rt != 0 && val[rt] != -1)
            {
                ans += val[rt];
                val[rt] = -1;
                rt = fail[rt];
            }
        }
        return ans;
    }*/
}ac;

char s[100100],t[350];

int main()
{
    Read();
    int n,val,i;
    while (~scanf("%s",s))
    {
        ac.Reset(); ac.Init();
        scanf("%d",&n);
        for (i = 0; i < n; ++i){
            scanf("%s %d",t,&val);
            ac.insert(t,val);
        }
        ac.Build();
        ac.find(s);
        //scanf("%s",tr);
        printf("%d\n",ac.dp[strlen(s)-1]);
    }
    return 0;
}