字符串水题，这么做可能比较巧妙。
1 /* 1462 */ 2 #include <iostream> 3 #include <string> 4 #include <map> 5 #include <queue> 6 #include <set> 7 #include <stack> 8 #include <vector> 9 #include <deque> 10 #include <algorithm> 11 #include <cstdio> 12 #include <cmath> 13 #include <ctime> 14 #include <cstring> 15 #include <climits> 16 #include <cctype> 17 #include <cassert> 18 #include <functional> 19 #include <iterator> 20 #include <iomanip> 21 using namespace std; 22 //#pragma comment(linker,"/STACK:102400000,1024000") 23 24 #define sti set<int> 25 #define stpii set<pair<int, int> > 26 #define mpii map<int,int> 27 #define vi vector<int> 28 #define pii pair<int,int> 29 #define vpii vector<pair<int,int> > 30 #define rep(i, a, n) for (int i=a;i<n;++i) 31 #define per(i, a, n) for (int i=n-1;i>=a;--i) 32 #define clr clear 33 #define pb push_back 34 #define mp make_pair 35 #define fir first 36 #define sec second 37 #define all(x) (x).begin(),(x).end() 38 #define SZ(x) ((int)(x).size()) 39 #define lson l, mid, rt<<1 40 #define rson mid+1, r, rt<<1|1 41 42 const int maxn = 30; 43 char M[maxn][maxn]; 44 char hs1[maxn], vs1[maxn], hs2[maxn], vs2[maxn]; 45 int ph1, pv1, ph2, pv2; 46 int hmark[26], vmark[26]; 47 48 bool f(char *hs, char *vs, int& ph, int& pv) { 49 int hlen = strlen(hs); 50 int vlen = strlen(vs); 51 52 memset(hmark, -1, sizeof(hmark)); 53 memset(vmark, -1, sizeof(vmark)); 54 rep(i, 0, hlen) { 55 if (hmark[hs[i]-'A'] < 0) 56 hmark[hs[i]-'A'] = i; 57 } 58 59 rep(i, 0, vlen) { 60 if (vmark[vs[i]-'A'] < 0) 61 vmark[vs[i]-'A'] = i; 62 } 63 64 rep(i, 0, hlen) { 65 if (vmark[hs[i]-'A'] >= 0) { 66 ph = i; 67 pv = vmark[hs[i]-'A']; 68 return true; 69 } 70 } 71 72 return false; 73 } 74 75 void solve() { 76 memset(M, ' ', sizeof(M)); 77 int hlen1 = strlen(hs1), hlen2 = strlen(hs2); 78 int vlen1 = strlen(vs1), vlen2 = strlen(vs2); 79 80 int c1 = ph1, c2 = ph2+hlen1+3; 81 int r = max(pv1, pv2); 82 int i, j, k = 0; 83 84 // vertical 85 int r1, r2; 86 int mx = 0; 87 88 if (pv1 >= pv2) { 89 r1 = 0; 90 r2 = pv1-pv2; 91 } else { 92 r2 = 0; 93 r1 = pv2-pv1; 94 } 95 96 for (i=0; i<vlen1; ++i,++r1) 97 M[r1][c1] = vs1[i]; 98 for (i=0; i<vlen2; ++i,++r2) 99 M[r2][c2] = vs2[i]; 100 mx = max(r1, r2); 101 102 // horizontal 103 for (i=0; i<hlen1; ++i) 104 M[r][k++] = hs1[i]; 105 k += 3; 106 for (i=0; i<hlen2; ++i) 107 M[r][k++] = hs2[i]; 108 109 rep(i, 0, mx) { 110 per(j, 0, maxn) { 111 if (M[i][j] != ' ') 112 break; 113 M[i][j] = ' '; 114 } 115 puts(M[i]); 116 } 117 } 118 119 int main() { 120 ios::sync_with_stdio(false); 121 #ifndef ONLINE_JUDGE 122 freopen("data.in", "r", stdin); 123 freopen("data.out", "w", stdout); 124 #endif 125 126 bool f1, f2; 127 int tt = 0; 128 129 while (scanf("%s",hs1)!=EOF && hs1[0]!='#') { 130 scanf("%s %s %s", vs1, hs2, vs2); 131 f1 = f(hs1, vs1, ph1, pv1); 132 f2 = f(hs2, vs2, ph2, pv2); 133 134 if (tt++) 135 putchar(' '); 136 if (f1 && f2) { 137 solve(); 138 } else { 139 puts("Unable to make two crosses"); 140 } 141 } 142 143 #ifndef ONLINE_JUDGE 144 printf("time = %d. ", (int)clock()); 145 #endif 146 147 return 0; 148 }