DP专辑
今天练了一波DP。时间紧迫我就只贴代码了。
20141120
fzu2129
http://acm.fzu.edu.cn/problem.php?pid=2129
不同的子序列个数
1 //#pragma comment(linker, "/STACK:102400000,102400000") 2 #include<cstdio> 3 #include<cmath> 4 #include<iostream> 5 #include<cstring> 6 #include<algorithm> 7 #include<cmath> 8 #include<map> 9 #include<set> 10 #include<stack> 11 #include<queue> 12 using namespace std; 13 #define mz(array) memset(array, 0, sizeof(array)) 14 #define mf1(array) memset(array, -1, sizeof(array)) 15 #define minf(array) memset(array, 0x3f, sizeof(array)) 16 #define REP(i,n) for(i=0;i<(n);i++) 17 #define FOR(i,x,n) for(i=(x);i<=(n);i++) 18 #define RD(x) scanf("%d",&x) 19 #define RD2(x,y) scanf("%d%d",&x,&y) 20 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z) 21 #define WN(x) printf("%d\n",x); 22 #define RE freopen("brackets.in","r",stdin) 23 #define WE freopen("brackets.out","w",stdout) 24 #define mp make_pair 25 #define pb push_back 26 #define pf push_front 27 #define ppf pop_front 28 #define ppb pop_back 29 typedef long long ll; 30 typedef unsigned long long ull; 31 32 const double pi=acos(-1.0); 33 const double eps=1e-10; 34 35 const ll MOD=1e9+7; 36 ll MMOD = MOD*1000000; 37 const int maxn=1111111; 38 39 int a[maxn]; 40 int n; 41 42 ll f[maxn]; 43 int d[maxn]; 44 45 ll farm() { 46 int i; 47 mf1(d); 48 mz(f); 49 f[0]=0; 50 FOR(i,1,n) { 51 if(d[a[i]]==-1) f[i]=(f[i-1] + f[i-1]+1)%MOD; 52 else{ 53 f[i]=f[i-1] + f[i-1] - f[d[a[i]]-1]; 54 f[i]+=MMOD; 55 f[i]%=MOD; 56 } 57 d[a[i]]=i; 58 } 59 return f[n]; 60 } 61 62 int main() { 63 //RE; 64 //WE; 65 int i; 66 while(RD(n)!=EOF) { 67 FOR(i,1,n)RD(a[i]); 68 printf("%I64d\n",farm()); 69 } 70 }
ASC17 A题
http://codeforces.com/gym/100221
不同的括号子序列个数，要用java的BigInteger。
1 import java.util.*; 2 import java.io.*; 3 import java.math.BigInteger; 4 5 public class Main { 6 public static final int MAXN = 333; 7 public static char[] s = null; 8 public static int N; 9 public static BigInteger[][] f = new BigInteger[MAXN][MAXN]; 10 public static final BigInteger F1 = BigInteger.valueOf(-1); 11 public static final BigInteger ZERO = BigInteger.valueOf(0); 12 public static final BigInteger ONE = BigInteger.valueOf(1); 13 14 public static BigInteger gank(int last, int o) { 15 // System.out.printf("Gank %d,%d\n",last,o); 16 if (last >= 0 && !f[last][o].equals(F1)) 17 return f[last][o]; 18 BigInteger re = ZERO; 19 if (o == 0) 20 re = ONE; 21 int i = last + 1; 22 while (i < N && (s[i] != '(')) 23 i++; 24 if (i < N) 25 re = re.add(gank(i, o + 1)); 26 if (o > 0) { 27 i = last + 1; 28 while (i < N && (s[i] != ')')) 29 i++; 30 if (i < N) 31 re = re.add(gank(i, o - 1)); 32 } 33 if (last >= 0) 34 f[last][o] = re; 35 return re; 36 } 37 38 public static void main(String[] args) throws Exception { 39 File fin = new File("brackets.in"); 40 File fout = new File("brackets.out"); 41 InputStream ins = new FileInputStream(fin); 42 OutputStream ous = new FileOutputStream(fout); 43 Reader.init(ins); 44 String S = Reader.next(); 45 s = S.toCharArray(); 46 N = S.length(); 47 for (int i = 0; i < MAXN; i++) { 48 for (int j = 0; j < MAXN; j++) { 49 f[i][j] = F1; 50 } 51 } 52 BigInteger ans = gank(-1, 0); 53 ous.write(ans.toString().getBytes()); 54 } 55 } 56 57 class Reader { 58 static BufferedReader reader; 59 static StringTokenizer tokenizer; 60 61 /** call this method to initialize reader for InputStream */ 62 static void init(InputStream input) { 63 reader = new BufferedReader(new InputStreamReader(input)); 64 tokenizer = new StringTokenizer(""); 65 } 66 67 /** get next word */ 68 static String next() throws IOException { 69 while (!tokenizer.hasMoreTokens()) { 70 // TODO add check for eof if necessary 71 tokenizer = new StringTokenizer(reader.readLine()); 72 } 73 return tokenizer.nextToken(); 74 } 75 76 static int nextInt() throws IOException { 77 return Integer.parseInt(next()); 78 } 79 80 static double nextDouble() throws IOException { 81 return Double.parseDouble(next()); 82 } 83 }
Knapsack
带路径输出的背包，感觉直接用二维数组会好一点，这个有点乱。
1 #include<cstdio> 2 #include<cmath> 3 #include<iostream> 4 #include<cstring> 5 #include<algorithm> 6 #include<cmath> 7 #include<map> 8 #include<set> 9 #include<stack> 10 #include<queue> 11 #include<cassert> 12 using namespace std; 13 #define mz(array) memset(array, 0, sizeof(array)) 14 #define mf1(array) memset(array, -1, sizeof(array)) 15 #define minf(array) memset(array, 0x3f, sizeof(array)) 16 #define REP(i,n) for(i=0;i<(n);i++) 17 #define FOR(i,x,n) for(i=(x);i<=(n);i++) 18 #define FORD(i,x,y) for(i=(x);i>=(y);i--) 19 #define RD(x) scanf("%d",&x) 20 #define RD2(x,y) scanf("%d%d",&x,&y) 21 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z) 22 #define WN(x) printf("%d\n",x); 23 #define RE freopen("knapsack.in","r",stdin) 24 #define WE freopen("knapsack.out","w",stdout) 25 #define mp make_pair 26 #define pb push_back 27 #define pf push_front 28 #define ppf pop_front 29 #define ppb pop_back 30 typedef long long ll; 31 typedef unsigned long long ull; 32 typedef long double LD; 33 34 const int maxm=20011; 35 const int maxn=2011; 36 37 int dp[maxm], c[maxm], n, m, w[maxn], v[maxn]; 38 int d[maxn][maxm]; 39 vector<int> V; 40 int main() { 41 RE; 42 WE; 43 // freopen("in.txt","r",stdin); 44 while(~scanf("%d%d", &n, &m)) { 45 memset(dp, 0 ,sizeof(dp)); 46 memset(c, -1, sizeof(c)); 47 mz(d); 48 assert(n>0&&n<=1000); 49 assert(m>0&&m<=10000); 50 for(int i=1; i<=n; i++) scanf("%d", &v[i]); 51 for(int i=1; i<=n; i++) scanf("%d", &w[i]); 52 for(int i=1;i<=n;i++){ 53 assert(v[i]>=0&&v[i]<=100); 54 assert(w[i]>=0&&w[i]<=100); 55 //while(1); 56 //return 0; 57 58 } 59 for(int i=1; i<=n; i++) { 60 for(int j=m; j>=v[i]; j--) { 61 assert(j>=v[i]); 62 if(dp[j] < dp[j - v[i]] + w[i]) { 63 dp[j] = dp[j - v[i]] + w[i]; 64 c[j] = i; 65 d[i][j]=c[j-v[i]]; 66 } 67 } 68 } 69 int tmp=0, mx = 0; 70 for(int i=m; i>0; i--) 71 if(dp[i] > mx) { 72 mx = dp[i]; 73 tmp = i; 74 } 75 V.clear(); 76 if(tmp==0){ 77 puts("0"); 78 continue; 79 } 80 V.pb(c[tmp]); 81 int i=d[c[tmp]][tmp]; 82 83 tmp-=v[c[tmp]]; 84 while(i > 0 && tmp>=0) { 85 V.pb(i); 86 int t=v[i]; 87 assert(tmp>=0&&tmp<=m); 88 i=d[i][tmp]; 89 tmp -= t; 90 } 91 int sz = V.size(); 92 printf("%d\n",sz); 93 sort(V.begin(),V.end()); 94 for(int i=0;i<sz;i++){ 95 printf("%d%c",V[i],i==sz-1?'\n':' '); 96 } 97 //if(sz==0)puts(""); 98 } 99 return 0; 100 }
Dominoes
15*100的1*2铺砖
1 //#pragma comment(linker, "/STACK:102400000,102400000") 2 #include<cstdio> 3 #include<cmath> 4 #include<iostream> 5 #include<cstring> 6 #include<algorithm> 7 #include<cmath> 8 #include<map> 9 #include<set> 10 #include<stack> 11 #include<queue> 12 #include<climits> 13 using namespace std; 14 #define mz(array) memset(array, 0, sizeof(array)) 15 #define mf1(array) memset(array, -1, sizeof(array)) 16 #define minf(array) memset(array, 0x3f, sizeof(array)) 17 #define REP(i,n) for(i=0;i<(n);i++) 18 #define FOR(i,x,n) for(i=(x);i<=(n);i++) 19 #define FORD(i,x,y) for(i=(x);i>=(y);i--) 20 #define RD(x) scanf("%d",&x) 21 #define RD2(x,y) scanf("%d%d",&x,&y) 22 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z) 23 #define WN(x) printf("%d\n",x); 24 #define RE freopen("dominoes.in","r",stdin) 25 #define WE freopen("dominoes.out","w",stdout) 26 #define mp make_pair 27 #define pb push_back 28 #define pf push_front 29 #define ppf pop_front 30 #define ppb pop_back 31 typedef long long ll; 32 typedef unsigned long long ull; 33 typedef long double LD; 34 const ll MOD=1e9; 35 int m,n; 36 37 const int maxs=(1<<15)+10; 38 const int maxn=111; 39 int f[maxn][maxs]; 40 41 inline void dfs(const int &H, const int &s,const int &ss,const int &d){ 42 if(d==m){ 43 f[H][ss] += f[H-1][s]; 44 f[H][ss] %= MOD; 45 return; 46 } 47 if(s&(1<<d))dfs(H,s,ss,d+1); 48 else{ 49 dfs(H,s, ss|(1<<d) ,d+1); 50 if(d<m-1 && ((s&(1<<(d+1)))==0)) 51 dfs(H, s, ss, d+2); 52 } 53 } 54 55 inline int farm(){ 56 int i,j,k; 57 int maxi=1<<m; 58 mz(f); 59 f[0][0]=1; 60 FOR(i,1,n){ 61 REP(j,maxi) 62 if(f[i-1][j]){ 63 dfs(i,j,0,0); 64 } 65 } 66 return f[n][0]; 67 } 68 69 int main(){ 70 RE; 71 WE; 72 while(RD2(m,n)!=EOF){ 73 printf("%d\n",farm()); 74 } 75 return 0; 76 }
Number of paths in acyclic graph
有向无环图判1~n的路径数。
1 //#pragma comment(linker, "/STACK:102400000,102400000") 2 #include<cstdio> 3 #include<cmath> 4 #include<iostream> 5 #include<cstring> 6 #include<algorithm> 7 #include<cmath> 8 #include<map> 9 #include<set> 10 #include<stack> 11 #include<queue> 12 using namespace std; 13 #define mz(array) memset(array, 0, sizeof(array)) 14 #define mf1(array) memset(array, -1, sizeof(array)) 15 #define minf(array) memset(array, 0x3f, sizeof(array)) 16 #define REP(i,n) for(i=0;i<(n);i++) 17 #define FOR(i,x,n) for(i=(x);i<=(n);i++) 18 #define FORD(i,x,y) for(i=(x);i>=(y);i--) 19 #define RD(x) scanf("%d",&x) 20 #define RD2(x,y) scanf("%d%d",&x,&y) 21 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z) 22 #define WN(x) printf("%d\n",x); 23 #define RE freopen("countpaths.in","r",stdin) 24 #define WE freopen("countpaths.out","w",stdout) 25 #define mp make_pair 26 #define pb push_back 27 #define pf push_front 28 #define ppf pop_front 29 #define maxn 100005 30 #define maxm 200005 31 32 int head[maxn]; 33 int en; 34 #define ll long long 35 struct pp{ 36 int v,next; 37 38 }e[maxm]; 39 void add(int u,int v){ 40 e[en].v=v; 41 e[en].next=head[u]; 42 head[u]=en++; 43 } 44 45 ll f[maxn]; 46 #define mod 1000000007 47 48 ll dfs(ll x){ 49 if(f[x]!=-1)return f[x]; 50 ll re=0; 51 int i; 52 for(i=head[x]; i!=-1; i=e[i].next){ 53 re+=dfs(e[i].v); 54 re%=mod; 55 } 56 f[x]=re; 57 return re; 58 } 59 60 int main(){ 61 int n,m; 62 RE; 63 WE; 64 scanf("%d%d",&n,&m); 65 memset(head,-1,sizeof(head)); 66 en=0; 67 for(int i=0;i<m;i++){ 68 int a,b; 69 scanf("%d%d",&a,&b); 70 add(a,b); 71 } 72 memset(f,-1,sizeof(f)); 73 f[n]=1; 74 cout<<dfs(1)<<endl; 75 }
Longest common subsequence
LCS带路径输出
1 //#pragma comment(linker, "/STACK:102400000,102400000") 2 #include<cstdio> 3 #include<cmath> 4 #include<iostream> 5 #include<cstring> 6 #include<algorithm> 7 #include<cmath> 8 #include<map> 9 #include<set> 10 #include<stack> 11 #include<queue> 12 using namespace std; 13 #define mz(array) memset(array, 0, sizeof(array)) 14 #define mf1(array) memset(array, -1, sizeof(array)) 15 #define minf(array) memset(array, 0x3f, sizeof(array)) 16 #define REP(i,n) for(i=0;i<(n);i++) 17 #define FOR(i,x,n) for(i=(x);i<=(n);i++) 18 #define FORD(i,x,y) for(i=(x);i>=(y);i--) 19 #define RD(x) scanf("%d",&x) 20 #define RD2(x,y) scanf("%d%d",&x,&y) 21 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z) 22 #define WN(x) printf("%d\n",x); 23 #define RE freopen("lcs.in","r",stdin) 24 #define WE freopen("lcs.out","w",stdout) 25 #define mp make_pair 26 #define pb push_back 27 #define pf push_front 28 #define ppf pop_front 29 #define ppb pop_back 30 typedef long long ll; 31 typedef unsigned long long ull; 32 typedef long double LD; 33 34 const int maxn=2111; 35 36 int n,m; 37 int a[maxn]; 38 int b[maxn]; 39 40 int f[maxn][maxn]; 41 pair<int,int> pre[maxn][maxn]; 42 43 int ans; 44 vector<int>an; 45 46 void farm() { 47 int i,j; 48 mz(f); 49 FOR(i,1,n) { 50 FOR(j,1,m) { 51 if(a[i]==b[j]) { 52 f[i][j]=f[i-1][j-1] + 1; 53 pre[i][j]=mp(i-1,j-1); 54 } else { 55 if(f[i-1][j]>f[i][j-1])f[i][j]=f[i-1][j] , pre[i][j]=mp(i-1,j); 56 else f[i][j]=f[i][j-1] , pre[i][j]=mp(i,j-1); 57 } 58 } 59 } 60 ans=f[n][m]; 61 i=n,j=m; 62 an.clear(); 63 while(f[i][j]!=0){ 64 int qi=pre[i][j].first; 65 int qj=pre[i][j].second; 66 if(f[i][j]>f[qi][qj])an.pb(a[i]); 67 i=qi,j=qj; 68 } 69 } 70 71 int main() { 72 RE; 73 WE; 74 int i; 75 while(RD(n)!=EOF) { 76 FOR(i,1,n)RD(a[i]); 77 RD(m); 78 FOR(i,1,m)RD(b[i]); 79 farm(); 80 printf("%d\n",ans); 81 if(ans>0)printf("%d",an[ans-1]); 82 FORD(i,ans-2,0){ 83 printf(" %d",an[i]); 84 } 85 puts(""); 86 } 87 return 0; 88 }
Levenshtein distance
编辑距离？队友写的，日后我再看看。
1 //#pragma comment(linker, "/STACK:102400000,102400000") 2 #include<cstdio> 3 #include<cmath> 4 #include<iostream> 5 #include<cstring> 6 #include<algorithm> 7 #include<cmath> 8 #include<map> 9 #include<set> 10 #include<stack> 11 #include<queue> 12 using namespace std; 13 #define mz(array) memset(array, 0, sizeof(array)) 14 #define mf1(array) memset(array, -1, sizeof(array)) 15 #define minf(array) memset(array, 0x3f, sizeof(array)) 16 #define REP(i,n) for(i=0;i<(n);i++) 17 #define FOR(i,x,n) for(i=(x);i<=(n);i++) 18 #define FORD(i,x,y) for(i=(x);i>=(y);i--) 19 #define RD(x) scanf("%d",&x) 20 #define RD2(x,y) scanf("%d%d",&x,&y) 21 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z) 22 #define WN(x) printf("%d\n",x); 23 #define RE freopen("levenshtein.in","r",stdin) 24 #define WE freopen("levenshtein.out","w",stdout) 25 #define mp make_pair 26 #define pb push_back 27 #define pf push_front 28 #define ppf pop_front 29 #define ppb pop_back 30 31 //#define inf 0x7fffffff 32 //char a[5005],b[5005]; 33 //int dp[5005][5005]; 34 //int main(){ 35 // while(scanf("%s%s",a,b)==2){ 36 // int len1=strlen(a); 37 // int len2=strlen(b); 38 // for(int i=0;i<=len1;i++) 39 // for(int j=0;j<=len2;j++)dp[i][j]=inf; 40 // for(int i=0;i<=len1;i++)dp[0][i]=i; 41 // for(int i=0;i<=len2;i++)dp[i][0]=i; 42 // for(int i=1;i<=len1;i++){ 43 // for(int j=1;j<=len2;j++){ 44 // dp[i][j]=min(dp[i-1][j]+1,min(dp[i][j-1]+1,dp[i-1][j-1]+(b[i]==a[j])?0:1)); 45 // } 46 // } 47 // cout<<dp[len1][len2]<<endl; 48 // } 49 // return 0; 50 //} 51 // 52 53 char s1[6000],s2[6000]; 54 int d[6000][6000]; 55 int min(int a,int b,int c) { 56 int t = a < b ? a : b; 57 return t < c ? t : c; 58 } 59 void editDistance(int len1,int len2) 60 { 61 int i,j; 62 for(i = 0;i <= len1;i++) 63 d[i][0] = i; 64 for(j = 0;j <= len2;j++) 65 d[0][j] = j; 66 for(i = 1;i <= len1;i++) 67 for(j = 1;j <= len2;j++) 68 { 69 int cost = s1[i-1] == s2[j-1] ? 0 : 1; 70 int deletion = d[i-1][j] + 1; 71 int insertion = d[i][j-1] + 1; 72 int substitution = d[i-1][j-1] + cost; 73 d[i][j] = min(deletion,insertion,substitution); 74 } 75 printf("%d\n",d[len1][len2]); 76 } 77 int main() 78 { 79 RE; 80 WE; 81 while(scanf("%s %s",s1,s2) != EOF) 82 editDistance(strlen(s1),strlen(s2)); 83 return 0; 84 }
Longest increasing subsequence
LIS，带路径输出
1 //#pragma comment(linker, "/STACK:102400000,102400000") 2 #include<cstdio> 3 #include<cmath> 4 #include<iostream> 5 #include<cstring> 6 #include<algorithm> 7 #include<cmath> 8 #include<map> 9 #include<set> 10 #include<stack> 11 #include<queue> 12 using namespace std; 13 #define mz(array) memset(array, 0, sizeof(array)) 14 #define mf1(array) memset(array, -1, sizeof(array)) 15 #define minf(array) memset(array, 0x3f, sizeof(array)) 16 #define REP(i,n) for(i=0;i<(n);i++) 17 #define FOR(i,x,n) for(i=(x);i<=(n);i++) 18 #define FORD(i,x,y) for(i=(x);i>=(y);i--) 19 #define RD(x) scanf("%d",&x) 20 #define RD2(x,y) scanf("%d%d",&x,&y) 21 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z) 22 #define WN(x) printf("%d\n",x); 23 #define RE freopen("lis.in","r",stdin) 24 #define WE freopen("lis.out","w",stdout) 25 #define mp make_pair 26 #define pb push_back 27 #define pf push_front 28 #define ppf pop_front 29 #define ppb pop_back 30 typedef long long ll; 31 typedef unsigned long long ull; 32 typedef long double LD; 33 34 const int maxn=5555; 35 36 int n; 37 int a[maxn]; 38 39 int f[maxn]; 40 int pre[maxn]; 41 42 int ans; 43 vector<int>an; 44 45 inline void farm() { 46 int i,j; 47 mz(pre); 48 FOR(i,1,n) { 49 f[i]=1; 50 FOR(j,1,i-1) { 51 if(a[i]>a[j] && f[j]+1>f[i]) { 52 f[i]=f[j]+1; 53 pre[i]=j; 54 } 55 } 56 } 57 ans=0; 58 int ai=0; 59 FOR(i,1,n) { 60 if(f[i]>ans)ans=f[i], ai=i; 61 } 62 i=ai; 63 an.clear(); 64 REP(j,ans) { 65 an.pb(a[i]); 66 i=pre[i]; 67 } 68 } 69 70 int main() { 71 RE; 72 WE; 73 int i; 74 while(RD(n)!=EOF) { 75 if(n>=maxn)while(1); 76 FOR(i,1,n)RD(a[i]); 77 farm(); 78 printf("%d\n",ans); 79 if(ans>0)printf("%d",an[ans-1]); 80 FORD(i,ans-2,0) { 81 printf(" %d",an[i]); 82 } 83 puts(""); 84 } 85 return 0; 86 }
Maximal weight matching in tree
树上选一些边，没有公共顶点，要求边权和最大。
树DP，队友写的。
哇，代码不见啦！反正就是f[i][j],j=0~1,0是不用这个点，1是用这个点。
Matrix multiplication
矩阵链乘，加括号使得代价最小，带路径输出。算法导论题，队友写的，碉
1 //#pragma comment(linker, "/STACK:102400000,102400000") 2 #include<cstdio> 3 #include<cmath> 4 #include<iostream> 5 #include<cstring> 6 #include<algorithm> 7 #include<cmath> 8 #include<map> 9 #include<set> 10 #include<stack> 11 #include<queue> 12 #include<climits> 13 using namespace std; 14 #define mz(array) memset(array, 0, sizeof(array)) 15 #define mf1(array) memset(array, -1, sizeof(array)) 16 #define minf(array) memset(array, 0x3f, sizeof(array)) 17 #define REP(i,n) for(i=0;i<(n);i++) 18 #define FOR(i,x,n) for(i=(x);i<=(n);i++) 19 #define FORD(i,x,y) for(i=(x);i>=(y);i--) 20 #define RD(x) scanf("%d",&x) 21 #define RD2(x,y) scanf("%d%d",&x,&y) 22 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z) 23 #define WN(x) printf("%d\n",x); 24 #define RE freopen("matrix.in","r",stdin) 25 #define WE freopen("matrix.out","w",stdout) 26 #define mp make_pair 27 #define pb push_back 28 #define pf push_front 29 #define ppf pop_front 30 #define ppb pop_back 31 typedef long long ll; 32 typedef unsigned long long ull; 33 typedef long double LD; 34 35 const int maxn=1000 + 10; 36 37 int n, a[maxn], b[maxn], m[maxn][maxn], s[maxn][maxn]; 38 void Print(int l ,int r){ 39 if(l == r){ 40 putchar('A'); 41 return; 42 } 43 putchar('('); 44 Print(l, s[l][r]); 45 Print(s[l][r] + 1, r); 46 putchar(')'); 47 } 48 int main(){ 49 RE; 50 WE; 51 scanf("%d", &n); 52 for(int i=1; i<=n; i++) scanf("%d%d", &a[i], &b[i]), m[i][i] = 0; 53 for(int l=2; l<=n; l++){ 54 for(int i=1; i<=n-l+1; i++){ 55 int j = i + l -1; 56 m[i][j] = INT_MAX; 57 for(int k=i; k<j; k++){ 58 int q = m[i][k] + m[k + 1][j] + a[i] * b[k] * b[j]; 59 if(q < m[i][j]){ 60 m[i][j] = q; 61 s[i][j] = k; 62 } 63 } 64 } 65 } 66 // cout << m[1][n] << endl; 67 Print(1, n); 68 return 0; 69 }
Longest subpalindrome
最长回文子串，带路径输出。
f[l][r]记录区间[l,r]的最长回文子串长度，区间从小到大搞。
1 //#pragma comment(linker, "/STACK:102400000,102400000") 2 #include<cstdio> 3 #include<cmath> 4 #include<iostream> 5 #include<cstring> 6 #include<algorithm> 7 #include<cmath> 8 #include<map> 9 #include<set> 10 #include<stack> 11 #include<queue> 12 #include<climits> 13 using namespace std; 14 #define mz(array) memset(array, 0, sizeof(array)) 15 #define mf1(array) memset(array, -1, sizeof(array)) 16 #define minf(array) memset(array, 0x3f, sizeof(array)) 17 #define REP(i,n) for(i=0;i<(n);i++) 18 #define FOR(i,x,n) for(i=(x);i<=(n);i++) 19 #define FORD(i,x,y) for(i=(x);i>=(y);i--) 20 #define RD(x) scanf("%d",&x) 21 #define RD2(x,y) scanf("%d%d",&x,&y) 22 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z) 23 #define WN(x) printf("%d\n",x); 24 #define RE freopen("palindrome.in","r",stdin) 25 #define WE freopen("palindrome.out","w",stdout) 26 #define mp make_pair 27 #define pb push_back 28 #define pf push_front 29 #define ppf pop_front 30 #define ppb pop_back 31 typedef long long ll; 32 typedef unsigned long long ull; 33 typedef long double LD; 34 35 const int maxn=2222; 36 37 char s[2222]; 38 int f[2222][2222]; 39 pair<int,int> pre[maxn][maxn]; 40 41 int ans; 42 vector<char>anl,anr; 43 44 void farm(){ 45 int len=strlen(s); 46 int i,j,k; 47 mz(f); 48 mf1(pre); 49 REP(i,len)f[i][i]=1 , pre[i][i]=mp(-1,-1); 50 FOR(k,1,len-1){ 51 FOR(i,0,len-2){ 52 j=i+k; 53 if(j>len-1)break; 54 if(s[i]==s[j]){ 55 if(j-i>1){ 56 f[i][j]=f[i+1][j-1]+2; 57 pre[i][j] = mp(i+1,j-1); 58 }else{ 59 f[i][j]=2; 60 pre[i][j]=mp(-1,-1); 61 } 62 }else{ 63 if(f[i][j-1] > f[i+1][j])f[i][j]=f[i][j-1] , pre[i][j]=mp(i,j-1); 64 else f[i][j]=f[i+1][j] , pre[i][j]=mp(i+1,j); 65 } 66 } 67 } 68 ans=f[0][len-1]; 69 i=0; 70 j=len-1; 71 anl.clear(); 72 anr.clear(); 73 int qi,qj; 74 while(i>=0 && j>=0){ 75 qi=pre[i][j].first; 76 qj=pre[i][j].second; 77 if(qi<0 || qj<0){ 78 if(f[i][j]!=0){ 79 if(i==j)anl.pb(s[i]); 80 else anl.pb(s[i]),anr.pb(s[i]); 81 } 82 break; 83 } 84 //printf("%d,%d %d,%d %d,%d\n",i,j,qi,qj,f[i][j] , f[qi][qj]); 85 if(f[i][j] > f[qi][qj]){ 86 if(i!=j)anl.pb(s[i]), anr.pb(s[i]); 87 else anl.pb(s[i]); 88 //printf("%d,%c\n",i!=j,s[i]); 89 } 90 i=qi; 91 j=qj; 92 } 93 } 94 95 int main(){ 96 int i; 97 RE; 98 WE; 99 while(scanf("%s",s)!=EOF){ 100 farm(); 101 printf("%d\n",ans); 102 REP(i,anl.size())putchar(anl[i]); 103 FORD(i,anr.size()-1,0)putchar(anr[i]); 104 puts(""); 105 } 106 }
Travelling salesman problem
旅行商问题，代码不见了。
f[i][j]，一个是状压的到过哪个地方的状态，另一个是最后在哪个点。
String Decomposition
把字符串弄成又几个子串增殖出来的
太难啦，还没做出来
Solid Tilings
另一种比较难的铺砖，还没看。
