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[];
void Init()
{
fail[] = ;
for (int i = ; i < Mc; ++i)
{
ID[i] = i;
}
}
void Reset()
{
CL(chd[],);
sz = ;
}
int idx(char c) {return c-'a';}
void insert(char *s,int key)
{
int p = ;
for (; *s; s++)
{
int k = ID[*s - 'a'];
if (!chd[p][k])
{
CL(chd[sz],);
val[sz] = ;
len[sz]=;
chd[p][k] = sz++;
}
len[chd[p][k]]=len[p]+;
p = chd[p][k];
}
val[p] =max(val[p], key);
}
void Build()
{
int *s = Q ,*e = Q,i;
for (i = ; i < Mc; ++i)
{
if (chd[][i])
{
*e++ = chd[][i];
fail[chd[][i]] = ;
}
}
while (s != e)
{
int u = *s++;
for (i = ; 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=;
for(int i=,temp,c;T[i];i++) {
dp[i]=;
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]<) //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[],t[]; int main()
{
Read();
int n,val,i;
while (~scanf("%s",s))
{
ac.Reset(); ac.Init();
scanf("%d",&n);
for (i = ; 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)-]);
}
return ;
}