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  • 【HDOJ】3386 Final Kichiku “Lanlanshu”

    数位DP。
    需要注意的是需要特殊处理前导0,另外连续的==匹配,不要计重了,尽量贪心的匹配掉。

      1 /* 3886 */
      2 #include <iostream>
      3 #include <sstream>
      4 #include <string>
      5 #include <map>
      6 #include <queue>
      7 #include <set>
      8 #include <stack>
      9 #include <vector>
     10 #include <deque>
     11 #include <bitset>
     12 #include <algorithm>
     13 #include <cstdio>
     14 #include <cmath>
     15 #include <ctime>
     16 #include <cstring>
     17 #include <climits>
     18 #include <cctype>
     19 #include <cassert>
     20 #include <functional>
     21 #include <iterator>
     22 #include <iomanip>
     23 using namespace std;
     24 //#pragma comment(linker,"/STACK:102400000,1024000")
     25 
     26 #define sti                set<int>
     27 #define stpii            set<pair<int, int> >
     28 #define mpii            map<int,int>
     29 #define vi                vector<int>
     30 #define pii                pair<int,int>
     31 #define vpii            vector<pair<int,int> >
     32 #define rep(i, a, n)     for (int i=a;i<n;++i)
     33 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
     34 #define clr                clear
     35 #define pb                 push_back
     36 #define mp                 make_pair
     37 #define fir                first
     38 #define sec                second
     39 #define all(x)             (x).begin(),(x).end()
     40 #define SZ(x)             ((int)(x).size())
     41 #define lson            l, mid, rt<<1
     42 #define rson            mid+1, r, rt<<1|1
     43 
     44 const int mod = 100000000;
     45 const int maxl = 105;
     46 const int maxn = 105;
     47 char ps[maxl], ps_[maxl];
     48 char sa[maxl], sb[maxl];
     49 int a[maxl], plen;
     50 int dp[maxl][maxn][2][10];
     51 bool dp_[maxl][maxn];
     52 
     53 void f(char *s, int& l) {
     54     int len = strlen(s);
     55     int i = 0;
     56 
     57     l = 0;
     58     while (i<len-1 && s[i]=='0')
     59         ++i;
     60     while (i < len) {
     61         s[l++] = s[i++]-'0';
     62     }
     63     s[l] = '';
     64 }
     65 
     66 bool check(char ch, int d, int dd) {
     67     if (ch == '/')
     68         return d < dd;
     69     if (ch == '-')
     70         return d == dd;
     71     if (ch == '\')
     72         return d > dd;
     73     return false;
     74 }
     75 
     76 bool judge(char *s, int len) {
     77     if (len==1 || check(ps[1], s[0], s[1])==false)
     78         return false;
     79 
     80     memset(dp_, false, sizeof(dp_));
     81     dp_[1][1] = true;
     82 
     83     rep(j, 1, plen) {
     84         rep(i, 1, len-1) {
     85             if (!dp_[j][i])
     86                 continue;
     87 
     88             if (check(ps[j], s[i], s[i+1]))
     89                 dp_[j][i+1] = true;
     90             if (check(ps[j+1], s[i], s[i+1]))
     91                 dp_[j+1][i+1] = true;
     92         }
     93     }
     94 
     95     return dp_[plen][len-1];
     96 }
     97 
     98 int cal(char *s, int len) {
     99     if (len <= 1)
    100         return 0;
    101 
    102     memset(dp, -1, sizeof(dp));
    103     rep(k, 0, s[0]) {
    104         rep(kk, 0, 10) {
    105             if (k == 0) {
    106                 if (dp[0][1][0][kk] == -1)
    107                     dp[0][1][0][kk] = 1;
    108                 else
    109                     ++dp[0][1][0][kk];
    110                 continue;
    111             }
    112 
    113             if (check(ps[1], k, kk)) {
    114                 if (dp[1][1][0][kk] == -1)
    115                     dp[1][1][0][kk] = 1;
    116                 else
    117                     ++dp[1][1][0][kk];
    118             }
    119         }
    120     }
    121     rep(kk, 0, s[1]+1) {
    122         int at = kk==s[1];
    123         if (check(ps[1], s[0], kk)) {
    124             if (dp[1][1][at][kk] == -1)
    125                 dp[1][1][at][kk] = 1;
    126             else
    127                 ++dp[1][1][at][kk];
    128         }
    129     }
    130 
    131     rep(i, 0, plen+1) {
    132         int ii = i + 1;
    133         rep(j, 1, len-1) {
    134             int jj = j + 1;
    135 
    136             // consider boundary
    137             if (dp[i][j][1][s[j]] >= 0) {
    138                 rep(k, 0, s[j+1]+1) {
    139                     int at = k==s[j+1];
    140                     
    141                     if (check(ps[ii], s[j], k)) {
    142                         if (dp[ii][jj][at][k] >= 0) {
    143                             dp[ii][jj][at][k] = (dp[ii][jj][at][k] + dp[i][j][1][s[j]]) % mod;
    144                         } else {
    145                             dp[ii][jj][at][k] = dp[i][j][1][s[j]];
    146                         }
    147                     } else if (check(ps[i], s[j], k)) {
    148                         if (dp[i][jj][at][k] >= 0) {
    149                             dp[i][jj][at][k] = (dp[i][jj][at][k] + dp[i][j][1][s[j]]) % mod;
    150                         } else {
    151                             dp[i][jj][at][k] = dp[i][j][1][s[j]];
    152                         }
    153                     }
    154                 }
    155             }
    156 
    157             // consider < boundary
    158             rep(k, 0, 10) {
    159                 if (dp[i][j][0][k] < 0)
    160                     continue;
    161                 rep(kk, 0, 10) {
    162                     if (i == 0) {
    163                         if (k == 0) {
    164                             if (dp[i][jj][0][kk] >= 0) {
    165                                 dp[i][jj][0][kk] = (dp[i][jj][0][kk] + dp[i][j][0][k]) % mod;
    166                             } else {
    167                                 dp[i][jj][0][kk] = dp[i][j][0][k];
    168                             }
    169                         } else {
    170                             if (check(ps[ii], k, kk)) {
    171                                 if (dp[ii][jj][0][kk] >= 0) {
    172                                     dp[ii][jj][0][kk] = (dp[ii][jj][0][kk] + dp[i][j][0][k]) % mod;
    173                                 } else {
    174                                     dp[ii][jj][0][kk] = dp[i][j][0][k];
    175                                 }
    176                             }
    177                         }
    178                         continue;
    179                     }
    180 
    181                     if (check(ps[ii], k, kk)) {
    182                         if (dp[ii][jj][0][kk] >= 0) {
    183                             dp[ii][jj][0][kk] = (dp[ii][jj][0][kk] + dp[i][j][0][k]) % mod;
    184                         } else {
    185                             dp[ii][jj][0][kk] = dp[i][j][0][k];
    186                         }
    187                     } else if (check(ps[i], k, kk)) {
    188                         if (dp[i][jj][0][kk] >= 0) {
    189                             dp[i][jj][0][kk] = (dp[i][jj][0][kk] + dp[i][j][0][k]) % mod;
    190                         } else {
    191                             dp[i][jj][0][kk] = dp[i][j][0][k];
    192                         }
    193                     }
    194                 }
    195             }
    196         }
    197     }
    198 
    199     int ret = 0;
    200 
    201     rep(k, 0, 10) {
    202         if (dp[plen][len-1][1][k] >= 0)
    203             ret = (ret + dp[plen][len-1][1][k]) % mod;
    204         if (dp[plen][len-1][0][k] >= 0)
    205             ret = (ret + dp[plen][len-1][0][k]) % mod;
    206     }
    207 
    208     return ret;
    209 }
    210 
    211 void solve() {
    212     int alen, blen;
    213     int ans = 0, tmp;
    214 
    215     plen = strlen(ps+1);
    216     f(sa, alen);
    217     f(sb, blen);
    218 
    219     tmp = cal(sb, blen);
    220     ans += tmp;
    221     tmp = cal(sa, alen);
    222     ans -= tmp;
    223     if (judge(sa, alen))
    224         ++ans;
    225 
    226     ans = (ans + mod) % mod;
    227     printf("%08d
    ", ans);
    228 }
    229 
    230 int main() {
    231     ios::sync_with_stdio(false);
    232     #ifndef ONLINE_JUDGE
    233         freopen("data.in", "r", stdin);
    234         freopen("data.out", "w", stdout);
    235     #endif
    236 
    237     while (scanf("%s", ps+1)!=EOF) {
    238         scanf("%s %s", sa, sb);
    239         solve();
    240     }
    241 
    242     #ifndef ONLINE_JUDGE
    243         printf("time = %d.
    ", (int)clock());
    244     #endif
    245 
    246     return 0;
    247 }
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  • 原文地址:https://www.cnblogs.com/bombe1013/p/5146364.html
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