HNU 13108-Just Another Knapsack Problem (ac自动机上的dp)

#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
#include<iostream>
using namespace std;
const int maxnode = 300010;
const int sigma_size = 26;
struct Trie{
    int dp[100010];
    int ch[maxnode][sigma_size];
    int val[maxnode];     //该单词在模式串中出现的次数
    int last[maxnode];
    int f[maxnode];         //失配数组
    int num[maxnode];     //该单词出现在文本串的次数
    int pre[maxnode];    //该单词的前驱
    int len[maxnode];    //以该单词结尾的单词长度
    int Char[maxnode];     //该单词对应的字母
    int road[maxnode];   //路径压缩优化 针对计算模式串出现的种数
    int sz;
    int Newnode()
    {
        val[sz] = f[sz] = last[sz] = len[sz] = num[sz] = 0;
        memset(ch[sz], 0, sizeof ch[sz]);
        return sz++;
    }
    void init(){
        sz=0;
        Newnode();
    }
    int idx(char c){ return c-‘a‘; }
    void build(char *s,int v){
        int u = 0;
        for(int i = 0, c; s[i] ;i++){
            c = idx(s[i]);
            if(!ch[u][c])
                ch[u][c] = Newnode();
            pre[ch[u][c]] = u;
            Char[ch[u][c]] = s[i];
            len[ch[u][c]] = len[u]+1;
            road[ch[u][c]] = 1;
            u = ch[u][c];
        }
        val[u] = max(val[u],v);
        //num[u] = 0;
        //return u;
    }
   void getFail(){
        queue<int> q;
        for(int i = 0; i<sigma_size; i++)
            if(ch[0][i]) q.push(ch[0][i]);
        int r, c, u, v;
        while(!q.empty()){
            r = q.front(); q.pop();
            for(c = 0; c<sigma_size; c++){
                u = ch[r][c];
                if(!u)continue;
                q.push(u);
                v = f[r];
                while(v && ch[v][c] == 0) v = f[v]; //沿失配边走上去 如果失配后有节点 且 其子节点c存在则结束循环
                f[u] = ch[v][c];
            }
        }
    }
    void find(char *T){
    //计算模式串出现的个数：（每种多次出现算多次）
        int j = 0,l=strlen(T);
        for(int i = 0, c, temp; i<l ; i++){
                dp[i]=0;
            c = idx(T[i]);
            while(j && ch[j][c]==0) j = f[j];
            j = ch[j][c];

            temp = j;
            while(temp){
                if(val[temp]){
                    if(i-len[temp]<0)
                        dp[i]=max(dp[i],val[temp]);
                    else if(dp[i-len[temp]])
                        dp[i]=max(dp[i],dp[i-len[temp]]+val[temp]);
                }
                temp = f[temp];
            }
        }
        printf("%d\n",dp[l-1]);
    }
    int find_kind(char *T){
    //计算种数， 重复出现的不再计算(若多个询问则要在此处加for(i=0->sz)lu[i]=1;
        int j = 0, i, c, temp,ans=0;
        for(i = 0; T[i]; i++){
            c = idx(T[i]);
            while(j && ch[j][c] == 0) j = f[j];
            j = ch[j][c];
            temp = j;
            while(temp && road[temp]){
                if(val[temp])
                {
                    ++ans;
                    val[temp] = 0;
                }
                road[temp] = 0;
                temp = f[temp];
            }
        }
        return ans;
    }
}ac;
int n,v;
char T[100100],P[310];
int main() {
    while(~scanf("%s",T)){
            ac.init();
        scanf("%d",&n);
        while(n--){
            scanf("%s%d",P,&v);
            ac.build(P,v);
        }
        ac.getFail();
        ac.find(T);
    }
}