Topcoder SRM 677 div2

SRM 677 div2
通过数：2
250：
问[L,R]区间内有多少个素数且它是回文数的数。
由于数据范围比较小，直接暴力就可以。

#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>using namespace std;class PalindromePrime {public:    int count(int, int);};const int MAXN = 1000 + 50;int prime[MAXN], cnt;int vis[MAXN];int a[10];bool parliment(int u){    int tcnt = 0;    while(u) a[++tcnt] = u % 10, u /= 10;    int head = 1, rear = tcnt;    while(head < rear){        if(a[head] != a[rear]) return false;        head++, rear--;    }    return true;}void init(){    for(int i = 1 ; i < MAXN ; i++) vis[i] = 0;    cnt = 0;    for(int i = 2 ; i < MAXN ; i++){        if(vis[i]) continue;        if(parliment(i))    prime[cnt++] = i;        for(int now = i ; now < MAXN ; now += i) vis[now] = 1;    }}int PalindromePrime::count(int L, int R) {    init();    int t1, t2;    t1 = t2 = 0;    while(t1 < cnt && prime[t1] < L) t1++;    while(t2 < cnt && prime[t2] <= R) t2++;    return t2 - t1;}<%:testing-code%>//Powered by [KawigiEdit] 2.0!

550：
四个字符串，可以合成一个字符串，这个字符串的连续子串中包含这四个字符串。
问这个合成的字符串最短多长。
先记录下四个字符串前后连接时相同的地方多少个就可以，手残调了一段时间。

#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>using namespace std;class FourStrings {public:    int shortestLength(string, string, string, string);};int len[15][15];string str[15];void solve_len(){    for(int i = 1 ; i < 5 ; i++){        for(int j = 1 ; j < 5 ; j++){            if(i == j) continue;            else{                len[i][j] = 0;                for(int t1 = 0 ; t1 < (int)str[i].size() ; t1++){                    int ok = 1;                    for(int t2 = t1 ; t2 - t1 < str[j].size() && t2 < str[i].size(); t2++){                        if(str[i][t2] != str[j][t2 - t1]){                            ok = 0;                            break;                        }                    }                    if(ok){                        len[i][j] = min(str[i].size() - t1 , str[j].size());                        break;                    }                }            }        }    }//    for(int i = 1 ; i <= 4 ; i++){//        for(int j = 1 ; j <= 4 ; j++) printf("%d ", len[i][j]);//        printf("\n");//    }//    printf("\n");}int vis[5];int number[5];int ans;void dfs(int u){    if(u == 5){        int temp = 0;        int pre = number[1];        for(int i = 2 ; i <= 4 ; i++){            temp += len[pre][number[i]];            if(len[pre][number[i]] < str[number[i]].size()) pre = number[i];        }//        if(temp == 4){//            for(int i = 1 ; i <= 4 ; i++) printf("%d ", number[i]);//        printf("\n");//        }        ans = max(ans, temp);    }    else{        for(int i = 1 ; i < 5 ; i++){            if(vis[i] == 0){                number[u] = i;                vis[i] = 1;                dfs(u + 1);                vis[i] = 0;            }        }    }}int FourStrings::shortestLength(string a, string b, string c, string d) {    str[1] = a;    str[2] = b;    str[3] = c;    str[4] = d;    solve_len();    for(int i = 1 ; i < 5 ; i++) vis[i] = 0;    ans = 0;    for(int i = 1 ; i < 5 ; i++){        if(vis[i] == 0){            vis[i] = 1;            number[1] = i;            dfs(2);            vis[i] = 0;        }    }    ans = a.size() + b.size() + c.size() + d.size() - ans;    return ans;}<%:testing-code%>//Powered by [KawigiEdit] 2.0!

900：
好题啊~
给n个点，起点0，终点1。现在给出边，每个边有一个英文字母编号。
每个点可以重复走，问从起点走到终点、最终走过的边连成的字符串是回文串的最短路径长度多少。
想的时候并没有什么思路，感觉数据范围很想状态压缩DP，但是由于每个点可以重复走就没有办法DP了。
实际上这个思路从开始就是错误的，因为把走的过程看成单纯的从一个点走到另外一个点。然而由于后面走的点没有办法计算，所以这样无法求解。
如果把问题转化一下，看成对于已经是回文串的一段来说，分别向左向右扩展两个相同字符的节点，那么这题就解决了。

#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>#include <algorithm>#include <cstring>#include <string>#include <vector>using namespace std;class PalindromePath {public:    int shortestLength(int, vector <int>, vector <int>, string);};const int MAXN = 20 + 3;int dp[5][MAXN][MAXN];int edge[MAXN][MAXN];vector<int>lin[MAXN];int PalindromePath::shortestLength(int n, vector <int> a, vector <int> b, string c) {    int flag = 0;    memset(edge, -1, sizeof(edge));    for(int i = 0 ; i < n ; i++) edge[i][i] = 1, lin[i].clear();    memset(dp, -1, sizeof(dp));    for(int i = 0 ; i < n ; i++) dp[flag][i][i] = 0;    for(int i = 0 ; i < (int)a.size() ; i++){        edge[a[i]][b[i]] = edge[b[i]][a[i]] = c[i] - 'a' + 1;        dp[!flag][a[i]][b[i]] = dp[!flag][b[i]][a[i]] = 1;        lin[a[i]].push_back(b[i]);        lin[b[i]].push_back(a[i]);    }//    for(int i = 1 ; i <= n ; i++)//    if(dp[flag][0][1] != -1) return dp[flag][0][1];    for(int len = 2 ; len <= (n * n) ; len++){        for(int x = 0 ; x < n ; x++){            for(int y = 0 ; y < n ; y++){                if(dp[flag][x][y] == len - 2){                    for(int i = 0 ; i < (int)lin[x].size() ; i++){                        for(int j = 0 ; j < (int)lin[y].size() ; j++){                            int u = lin[x][i];                            int v = lin[y][j];                            if(edge[u][x] == edge[v][y]) dp[(flag + 2) % 3][u][v] = len;                        }                    }                }            }        }        if(dp[flag][0][1] != -1) return dp[flag][0][1];        flag = (flag + 1) % 3;    }    int ans = MAXN*MAXN + 1;    for(int i = 0 ; i < 4 ; i++) if(dp[i][0][1] != -1) ans = min(ans, dp[i][0][1]);    if(ans == MAXN * MAXN + 1) return -1;    else return ans;}<%:testing-code%>//Powered by [KawigiEdit] 2.0!