• hdu 1979 DFS + 字典树剪枝


    http://acm.hdu.edu.cn/showproblem.php?pid=1979

    Fill the blanks

    Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 373    Accepted Submission(s): 155


    Problem Description
    There is a matrix of 4*4, you should fill it with digits 0 – 9, and you should follow the rules in the following picture:

     
    Input
    No input.
     
    Output
    Print all the matrixs that fits the rules in the picture. 
    And there is a blank line between the every two matrixs.
     
    Sample Output
    1193 1009 9221 3191
    1193 1021 9029 3911 ……
    9173 1559 3821 3391
     
    Author
    8600
     
    Source
     

    1000 --- 9999中有204个顺着和倒着读都是素数的数。

    考虑的就是暴力dfs,然后最后再判断?超时。可以打表。

    可以用字典树维护前缀,

    每次都维护主对角线和副对角线的数字,还有四条列。然后如果不存在这样的前缀,直接剪掉就好。

    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #define IOS ios::sync_with_stdio(false)
    using namespace std;
    #define inf (0x3f3f3f3f)
    typedef long long int LL;
    #define MY "H:/CodeBlocks/project/CompareTwoFile/DataMy.txt", "w", stdout
    #define ANS "H:/CodeBlocks/project/CompareTwoFile/DataAns.txt", "w", stdout
    
    
    #include <iostream>
    #include <sstream>
    #include <vector>
    #include <set>
    #include <map>
    #include <queue>
    #include <string>
    const int maxn=1e5+20;
    bool prime[maxn];//这个用bool就够了,
    bool check[maxn];
    int goodprime[maxn];
    char strprime[204 + 2][10];
    int lenprime = 0;
    struct node {
        int cnt;
        struct node * pNext[10];
    } tree[maxn], *T;
    int num;
    struct node * create() {
        struct node * p = &tree[num++];
        for (int i = 0; i <= 9; ++i) {
            p->pNext[i] = NULL;
        }
        p->cnt = 1;
        return p;
    }
    void insert(struct node **T, int val) {
        struct node *p = *T;
        if (p == NULL) {
            *T = p = create();
        }
        char str[11] = {0};
        int lenstr = 0;
        while (val / 10 > 0) {
            str[++lenstr] = val % 10 + '0';
            val /= 10;
        }
        str[++lenstr] = val + '0';
        str[lenstr + 1] = '';
        strcpy(strprime[lenprime] + 1, str + 1);
    //    printf("%s
    ", str + 1);
        for (int i = 1; str[i]; ++i) {
            int id = str[i] - '0';
            if (p->pNext[id]) {
                p->pNext[id]->cnt++;
            } else p->pNext[id] = create();
            p = p->pNext[id];
        }
        return;
    }
    int find(struct node *T, int val) {
        struct node *p = T;
        if (!p) return 0;
        char str[11] = {0};
        int lenstr = 0;
        while (val / 10 > 0) {
            str[++lenstr] = val % 10 + '0';
            val /= 10;
        }
        str[++lenstr] = val + '0';
        str[lenstr + 1] = '';
        reverse(str + 1, str + 1 + lenstr);
    //    printf("%s
    ", str + 1);
        for (int i = 1; str[i]; ++i) {
            int id = str[i] - '0';
            if (!p->pNext[id]) return 0;
            p = p->pNext[id];
        }
        return p->cnt;
    }
    void init_prime() {
        for (int i = 2; i <= maxn - 20; i++) {
            if (!check[i]) { //说明i是质数
                prime[i] = true;
                for (int j = 2 * i; j <= maxn - 20; j += i) { //筛掉i的倍数
                    check[j] = true; //那么j就没可能是质数了
                    //book[j]=i; //表示j的最大质因数是i,不断更新。后面的质因数更大
                    //用这个的时候,需要把2*i变成i,否则book[2]不行。
                }
            }
        }
        for (int i = 1000; i <= 9999; ++i) {
            int t = 0;
            int h = i;
            if (!prime[i]) continue;
            while (h / 10 > 0) {
                t = t * 10 + h % 10;
                h /= 10;
            }
            t = t * 10 + h;
            if (prime[t]) {
                goodprime[++lenprime] = i;
                insert(&T, t);
            }
        }
        return ;
    }
    int f[66];
    bool book[204 + 20];
    int ans;
    bool tocheck(int toval[]) {
        for (int i = 1; i <= 4; ++i) {
            if (!find(T, toval[i])) return false;
        }
        return true;
    }
    void dfs(int cur, int valmain, int valother, int toval[]) {
        if (cur == 5) {
            ++ans;
            for (int i = 1; i <= 4; ++i) {
                printf("%d
    ", goodprime[f[i]]);
            }
            if (ans != 136) printf("
    ");
    //        while(1);
            return;
        }
        for (int i = 1; i <= lenprime; ++i) {
            f[cur] = i;
            if (cur == 1) {
                valmain = valmain * 10 + strprime[i][1] - '0';
                valother = valother * 10 + strprime[i][4] - '0';
            } else if (cur == 2) {
                valmain = valmain * 10 + strprime[i][2] - '0';
                valother = valother * 10 + strprime[i][3] - '0';
            } else if (cur == 3) {
                valmain = valmain * 10 + strprime[i][3] - '0';
                valother = valother * 10 + strprime[i][2] - '0';
            } else {
                valmain = valmain * 10 + strprime[i][4] - '0';
                valother = valother * 10 + strprime[i][1] - '0';
            }
            for (int j = 1; j <= 4; ++j) {
                toval[j] = toval[j] * 10 + strprime[i][j] - '0';
            }
            if (!find(T, valmain) || !find(T, valother) || !tocheck(toval)) {
                valmain /= 10;
                valother /= 10;
                for (int j = 1; j <= 4; ++j) {
                    toval[j] /= 10;
                }
                continue;
            }
    //        printf("%d
    ", ++ans);
            dfs(cur + 1, valmain, valother, toval);
            valmain /= 10;
            valother /= 10;
            for (int j = 1; j <= 4; ++j) {
                toval[j] /= 10;
            }
        }
    }
    void work() {
    //    printf("%d
    ", prime[9029]);
    //    printf("%d
    ", find(T, 9209));
        int toval[50] = {0};
        dfs(1, 0, 0, toval);
    //    cout << ans << endl;
    }
    
    int main() {
    #ifdef local
        freopen("data.txt","r",stdin);
    #endif
        init_prime();
        work();
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/liuweimingcprogram/p/6103216.html
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