HDU 4681 String dp

来源：互联网发布：大数据产业联盟编辑：程序博客网时间：2024/04/29 12:49

????:??
??:
??a,b,c?,??????????d?

??d??ab?????,??c??d????

????d???????

??:

????1000,???????????a????????b??????

????ansL + lenc + ansR

ansL??ab???????????,ansR?????????????????

lenc????????????????

???????n^2?????????

???????????

//HDU 4681 O(n^2)#include <bits/stdc++.h>using namespace std;const int maxn = 1050;int dp1[maxn][maxn], dp2[maxn][maxn];int l1[maxn], l2[maxn], len1, len2, len3;char s1[maxn], s2[maxn], s3[maxn];void update(int &x, int y){x = max(x, y);}void pre_deal(){    for(int i = 1; i <= len1; i++){        int l = 1;        for(int j = i; j <= len1; j++){            if(s1[j] == s3[l]) l++;            if(l > len3){                l1[i] = j;                break;            }        }    }    for(int i = 1; i <= len2; i++){        int l = 1;        for(int j = i; j <= len2; j++){            if(s2[j] == s3[l]) l++;            if(l > len3){                l2[i] = j;                break;            }        }    }    for(int i  = 1; i <= len1; i++){        for(int j = 1; j <= len2; j++){            if(s1[i] == s2[j]){                update(dp1[i][j], dp1[i-1][j-1]+1);            }            update(dp1[i][j], dp1[i-1][j]);            update(dp1[i][j], dp1[i][j-1]);        }    }    reverse(s1+1, s1+len1+1);    reverse(s2+1, s2+len2+1);    for(int i = 1; i <= len1; i++){        for(int j = 1; j <= len2; j++){            if(s1[i] == s2[j]){                update(dp2[i][j], dp2[i-1][j-1]+1);            }            update(dp2[i][j], dp2[i-1][j]);            update(dp2[i][j], dp2[i][j-1]);        }    }}int main(){    int T, ks = 0;    scanf("%d", &T);    while(T--){        memset(dp1, 0, sizeof(dp1));        memset(dp2, 0, sizeof(dp2));        memset(l1, -1, sizeof(l1));        memset(l2, -1, sizeof(l2));        scanf("%s", s1+1);        scanf("%s", s2+1);        scanf("%s", s3+1);        len1 = strlen(s1+1);        len2 = strlen(s2+1);        len3 = strlen(s3+1);        pre_deal();        int ans = 0;        for(int i = 1; i <= len1; i++){            for(int j = 1; j <= len2; j++){                if(l1[i] == -1 || l2[j] == -1) continue;                update(ans, dp1[i-1][j-1]+len3+dp2[len1-l1[i]][len2-l2[j]]);            }        }        printf("Case #%d: %d\n", ++ks, ans);    }    return 0;}

0 0