本文共 2316 字，大约阅读时间需要 7 分钟。
题目链接：
题意：给出一个串。星号可以被任意替换（包括空串），问号必须且只能被一个字母替换，感叹号必须且只能被三个字符替换。求替换后的最短回文串？最短的情况下字典序最小？
思路：f[i][j]表示i到j形成回文串的最小。长度从小到大枚举即可。
#include #include #include #include #include #include #include #include #include #include #include #define max(x,y) ((x)>(y)?(x):(y))#define min(x,y) ((x)<(y)?(x):(y))#define abs(x) ((x)>=0?(x):-(x))#define i64 long long#define u32 unsigned int#define u64 unsigned long long#define clr(x,y) memset(x,y,sizeof(x))#define CLR(x) x.clear()#define ph(x) push(x)#define pb(x) push_back(x)#define Len(x) x.length()#define SZ(x) x.size()#define PI acos(-1.0)#define sqr(x) ((x)*(x))#define MP(x,y) make_pair(x,y)#define FOR0(i,x) for(i=0;i =0;i--)#define DOW1(i,x) for(i=x;i>=1;i--)#define DOW(i,a,b) for(i=a;i>=b;i--)using namespace std;void RD(int &x){scanf("%d",&x);}void RD(i64 &x){scanf("%I64d",&x);}void RD(u32 &x){scanf("%u",&x);}void RD(double &x){scanf("%lf",&x);}void RD(int &x,int &y){scanf("%d%d",&x,&y);}void RD(i64 &x,i64 &y){scanf("%I64d%I64d",&x,&y);}void RD(u32 &x,u32 &y){scanf("%u%u",&x,&y);}void RD(double &x,double &y){scanf("%lf%lf",&x,&y);}void RD(int &x,int &y,int &z){scanf("%d%d%d",&x,&y,&z);}void RD(i64 &x,i64 &y,i64 &z){scanf("%I64d%I64d%I64d",&x,&y,&z);}void RD(u32 &x,u32 &y,u32 &z){scanf("%u%u%u",&x,&y,&z);}void RD(double &x,double &y,double &z){scanf("%lf%lf%lf",&x,&y,&z);}void RD(char &x){x=getchar();}void RD(char *s){scanf("%s",s);}void RD(string &s){cin>>s;}void PR(int x) {printf("%d\n",x);}void PR(i64 x) {printf("%I64d\n",x);}void PR(u32 x) {printf("%u\n",x);}void PR(u64 x) {printf("%llu\n",x);}void PR(double x) {printf("%.4lf\n",x);}void PR(char x) {printf("%c\n",x);}void PR(char *x) {printf("%s\n",x);}void PR(string x) {cout< < L||f[p2][i]==L&&strcmp(g[p2][i],s)>0) { assign(s,L); }}int main(){ gets(b); for(i=0;b[i];i++) { if(b[i]=='!') a[n++]='?',a[n++]='?',a[n++]='?'; else a[n++]=b[i]; } for(i=0;i =800||c1!=c2&&c1!='?'&&c2!='?') { assign("",800); } else { if(c1=='?') { if(c2=='?') c1='a'; else c1=c2; } strcpy(g[p2][i]+1,g[p0][i+1]); k=f[p0][i+1]; g[p2][i][0]=g[p2][i][k+1]=c1; g[p2][i][k+2]=0; f[p2][i]=k+2; } } } p0=p1; p1=p2; p2=3^p0^p1; } if(f[p1][0]<800) { puts("YES"); puts(g[p1][0]); } else { puts("NO"); } return 0;}
转载地址：http://whkjl.baihongyu.com/