【HDOJ】1462 Word Crosses

字符串水题，这么做可能比较巧妙。

 /* 1462 */
 #include <iostream>
 #include <string>
 #include <map>
 #include <queue>
 #include <set>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cassert>
 #include <functional>
 #include <iterator>
 #include <iomanip>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,1024000")

 #define sti                set<int>
 #define stpii            set<pair<int, int> >
 #define mpii            map<int,int>
 #define vi                vector<int>
 #define pii                pair<int,int>
 #define vpii            vector<pair<int,int> >
 #define rep(i, a, n)     for (int i=a;i<n;++i)
 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 #define clr                clear
 #define pb                 push_back
 #define mp                 make_pair
 #define fir                first
 #define sec                second
 #define all(x)             (x).begin(),(x).end()
 #define SZ(x)             ((int)(x).size())
 #define lson            l, mid, rt<<1
 #define rson            mid+1, r, rt<<1|1

 const int maxn = ;
 char M[maxn][maxn];
 char hs1[maxn], vs1[maxn], hs2[maxn], vs2[maxn];
 int ph1, pv1, ph2, pv2;
 int hmark[], vmark[];

 bool f(char *hs, char *vs, int& ph, int& pv) {
     int hlen = strlen(hs);
     int vlen = strlen(vs);

     memset(hmark, -, sizeof(hmark));
     memset(vmark, -, sizeof(vmark));
     rep(i, , hlen) {
         if (hmark[hs[i]-'A'] < )
             hmark[hs[i]-'A'] = i;
     }

     rep(i, , vlen) {
         if (vmark[vs[i]-'A'] < )
             vmark[vs[i]-'A'] = i;
     }

     rep(i, , hlen) {
         if (vmark[hs[i]-'A'] >= ) {
             ph = i;
             pv = vmark[hs[i]-'A'];
             return true;
         }
     }

     return false;
 }

 void solve() {
     memset(M, ' ', sizeof(M));
     int hlen1 = strlen(hs1), hlen2 = strlen(hs2);
     int vlen1 = strlen(vs1), vlen2 = strlen(vs2);

     int c1 = ph1, c2 = ph2+hlen1+;
     int r = max(pv1, pv2);
     int i, j, k = ;

     // vertical
     int r1, r2;
     int mx = ;

     if (pv1 >= pv2) {
         r1 = ;
         r2 = pv1-pv2;
     } else {
         r2 = ;
         r1 = pv2-pv1;
     }

     for (i=; i<vlen1; ++i,++r1)
         M[r1][c1] = vs1[i];
     for (i=; i<vlen2; ++i,++r2)
         M[r2][c2] = vs2[i];
     mx = max(r1, r2);

     // horizontal
     for (i=; i<hlen1; ++i)
         M[r][k++] = hs1[i];
     k += ;
     for (i=; i<hlen2; ++i)
         M[r][k++] = hs2[i];

     rep(i, , mx) {
         per(j, , maxn) {
             if (M[i][j] != ' ')
                 break;
             M[i][j] = '\0';
         }
         puts(M[i]);
     }
 }

 int main() {
     ios::sync_with_stdio(false);
     #ifndef ONLINE_JUDGE
         freopen("data.in", "r", stdin);
         freopen("data.out", "w", stdout);
     #endif

     bool f1, f2;
     int tt = ;

     while (scanf("%s",hs1)!=EOF && hs1[]!='#') {
         scanf("%s %s %s", vs1, hs2, vs2);
         f1 = f(hs1, vs1, ph1, pv1);
         f2 = f(hs2, vs2, ph2, pv2);

         if (tt++)
             putchar('\n');
         if (f1 && f2) {
             solve();
         } else {
             puts("Unable to make two crosses");
         }
     }

     #ifndef ONLINE_JUDGE
         printf("time = %d.\n", (int)clock());
     #endif

     return ;
 }