light oj 1073 状态压缩dp+输出字典序最小的解
算是比较基本的题，不忍直视的搓代码，string乱搞后惨不忍睹，调了好久终于AC，代码能力弱成渣。
做法：给你的一些串中，你先暴力去掉一些无用的子串， 比方有aaaa 和aa，那aa就没用了，去掉
然后从下往上放字符串（注意方向，为了输出字典序最小），我的val[i][j]表示串i在下，串j在上所增加的str[i]的前缀长度 如 aabb bbcccc， 那前缀为aa，长度为2
然后dfs搜出解。string用得太逗比了，这里用char字符串也还可以，写得好就很方便。
附上刚刚AC的代码和以前AC的代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
using namespace std;
string str[17];
int n;
bool ok(const string &a, const string &b) {
int sz = max<int>(0, b.size()-a.size());
int i, j, cur;
for(i = 0; i < sz; i++) {
cur = i;
for(j = 0; j < (int)a.size() &&
cur < (int)b.size()&& a[j] == b[cur]; j++, cur++);
if(j == (int)a.size()) return 1;
}
return 0;
}
int cal(const string &a , const string &b) {
int i, j, cur;
int ret = b.size();
for(i = 0; i < (int)b.size(); i++) {
cur = i;
for(j = 0; j < (int)a.size() &&
cur < (int)b.size() && a[j] == b[cur]; j++, cur++);
if(cur == (int)b.size())
ret = min<int>(ret, i);
}
return ret;
}
int val[17][17];
int dp[17][1<<17];
void out(int pos, int st) {
if(st == (st&-st)) {
//cout << "nice" << endl;
cout << str[pos];
return;
}
int j, cur = -1;
string tp = "Z";
int ans = -1;
for(j = 0; j < n; j++) if(j != pos && (st&(1<<j))) {
// cout << "j " << j << endl;
if(dp[j][st^(1<<pos)] + val[j][pos] == dp[pos][st]) {
// cout << "first " << str[pos].size()-val[j][pos];
string res = str[j].substr(str[pos].size()-val[j][pos], str[j].size());
// cout << "res~~~~" << res << endl;
if(res < tp) {
tp = res;
cur = j;
ans = val[j][pos];
}
}
}
// cout << "ans = " << ans << endl;
//cout << "~~~~~~~~~~~~~" << endl;
cout << str[pos].substr(0, val[cur][pos]);
//cout << "cur~~~" <<cur<< endl;
out(cur, st^(1<<pos));
}
int full;
int main() {
int i, j, k, cas, ca = 1;
cin >> cas;
while(cas--) {
cin >> n;
for(i = 0; i < n; i++)
cin >> str[i];
int m = 0;
for(i = 0; i < n; i++) {
for(j = 0; j < n; j++)if(j != i)
if(ok(str[i], str[j])) break;
if(j == n)
str[m++] = str[i];
}
n = m;
full = (1<<n)-1;
// for(i = 0; i < n; i++)
// cout << str[i] << endl;
for(i = 0; i < n; i++)
for(j = 0; j < n; j++) if(i != j) {
val[i][j] = cal(str[i], str[j]);
// printf("val[%d][%d] = %d\n", i, j, val[i][j]);
}
for(i = 0; i < n; i++)
for(j = 0; j < (1<<n); j++)
dp[i][j] = -1;
for(i = 0; i < n; i++)
dp[i][1<<i] = str[i].size();
for(j = 0; j < (1<<n); j++)
for(i = 0; i < n; i++) if(~dp[i][j]) {
for(k = 0; k < n; k++) if(!(j&(1<<k))) {
int inc = dp[i][j] + val[i][k];
//cout << inc << endl;
if(dp[k][j|(1<<k)] == -1 || dp[k][j|(1<<k)] > inc) {
dp[k][j|(1<<k)] = inc;
}
}
}
//for(i = 0; i < n; i++)
// cout << "i = " << dp[i][full] << endl;
int pos = 0;
// cout << (str[1] < str[2]) << endl;
// cout << n <<"~~~" << endl;
for(i = 1; i < n; i++)
if( dp[i][full] < dp[pos][full] ||
(dp[i][full] == dp[pos][full] && str[i] < str[pos])) {
pos = i;
// cout << "update" << pos << endl;
}
cout << "Case " << ca++ << ": ";
// cout << "pos = " << pos << endl;
// cout << "~~~" << dp[pos][full] << endl;
//cout << full << endl;
out(pos, full);
cout << endl;
}
return 0;
}
/*
1
10
TTTT
TT
T
T
T
GCGCG
CCCC
C
C
C
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int inf = 1e9;
struct node {
int len;
char s[103];
void in() {
scanf("%s", s);
len = strlen(s);
}
}p[16];
int n;
int cnt[16][16];
int two[18];
int dp[15][33333];
int Log[33333];
int f[101];
inline void getfail(char *s, int &la) {
int i, j = 0;
f[0] = f[1] = 0;
for(i = 1; i < la; i++) {
while(j && s[i] != s[j]) j = f[j];
if(s[i] == s[j]) j++;
f[i+1] = j;
}
}
inline bool kmp(char *s, char *t, int &la, int &lb) {
int i, j = 0;
getfail(t, lb);
for(i = 0; i < la; i++) {
while(j && s[i] != t[j]) j = f[j];
if(s[i] == t[j]) j++;
if(j == lb) return 1;
}
return 0;
}
inline bool ok(int &x, int &y) {
int i, j;
for(i = 0; i < p[x].len; i++) {
for(j = 0; j < p[y].len && i+j < p[x].len && p[x].s[i+j] == p[y].s[j]; j++);
if(j == p[y].len) return 1;
}
return 0;
}
bool cmp1(const node &a, const node &b) {
return a.len > b.len;
}
bool cmp2(const node &a, const node &b) {
return strcmp(a.s, b.s) < 0;
}
inline void out(int pos, int st) {
if(!st) {
puts(p[pos].s);
return;
}
int i;
int ans = -1, idx;
for(int x = ~two[n]&st; x; x -= x&-x) {
i = Log[x&-x];
int inc = dp[i][st^two[i]] + p[pos].len - cnt[pos][i];
if(ans == -1 || ans > inc || (ans == inc && strcmp(&p[i].s[cnt[pos][i]],&p[idx].s[cnt[pos][idx]]) < 0) ) {
ans = inc;
idx = i;
}
}
int end = p[pos].len - cnt[pos][idx];
for(i = 0; i < end; i++)
printf("%c", p[pos].s[i]);
out(idx, st^two[idx]);
}
int main() {
int i, j, x, y, cas;
two[0] = 1; Log[0] = -1;
for(i = 1; i <= 16; i++) two[i] = two[i-1] << 1;
for(i = 1; i < two[15]; i++) Log[i] = Log[i>>1] + 1;
scanf("%d", &cas);
for(int ca = 1; ca <= cas; ca++) {
scanf("%d", &n);
for(i = 0; i < n; i++) p[i].in();
sort(p, p+n, cmp1);
int m = 1;
for(i = 1; i < n; i++) {
getfail(p[i].s, p[i].len);
for(j = 0; j < m; j++) if(kmp(p[j].s, p[i].s, p[j].len, p[i].len)) break;
if(j == m) {
p[m++] = p[i];
}
}
n = m;
sort(p, p+n, cmp2);
for(i = 0; i < n; i++) {
for(j = 0; j < n; j++) if(i != j){
cnt[i][j] = 0;
for(x = 0; x < p[i].len; x++) {
int tp = 0;
for(y = 0; y < p[j].len && x+y < p[i].len && p[i].s[x+y] == p[j].s[y]; y++) tp++;
if(x+y == p[i].len) cnt[i][j] = max(cnt[i][j], tp);
}
}
}
printf("Case %d: ", ca);
for(i = 0; i < n; i++)
for(j = 0; j < two[n]; j++)
dp[i][j] = -1;
int ans = -1, pos;
for(i = 0; i < n; i++)
dp[i][0] = p[i].len;
for(j = 0; j < two[n]; j++) {
for(i = 0; i < n; i++) if(~dp[i][j]){
for(int xx = (two[n+1]-1)^(~two[n]&j); xx; xx -= xx&-xx) {
x = Log[xx&-xx];
if(i == x) continue;
int inc = dp[i][j] + p[x].len - cnt[x][i];
if(dp[x][j|two[i]] == -1 || dp[x][j|two[i]] > inc) {
dp[x][j|two[i]] = inc;
}
}
}
}
for(i = 0; i < n; i++) {
if(ans == -1 || ans > dp[i][(two[n]-1)^two[i]]) {
ans = dp[i][(two[n]-1)^two[i]];
pos = i;
}
}
out(pos, (two[n]-1)^two[pos]);
}
}