标签：can 匹配 cto div tac 1.0 inf putchar return
给出两个长度小于1000的字符串，有三种操作，插入一个字符，删除一个字符，替换一个字符。
问A变成B所需的最少操作数(即编辑距离)
考虑DP，可以用反证法证明依次从头到尾对A，B进行匹配是不会影响答案的
令dp[i][j]表示A[i]~[lenA]变成B[j]~[lenB]的最优解。
如果把B[j]插入到A[i]前，dp[i][j]=dp[i][j+1]+1
如果删除A[i]，dp[i][j]=dp[i+1][j]+1.
如果A[i]==B[j], dp[i][j]=dp[i+1][j+1].
如果A[i]!=B[j], 则用B[j]替换A[i]. dp[i][j]=dp[i+1][j+1]+1.
得出状态转移方程后记搜一下就OK了。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-3 # define MOD 100000007 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int res=0, flag=0; char ch; if((ch=getchar())==‘-‘) flag=1; else if(ch>=‘0‘&&ch<=‘9‘) res=ch-‘0‘; while((ch=getchar())>=‘0‘&&ch<=‘9‘) res=res*10+(ch-‘0‘); return flag?-res:res; } void Out(int a) { if(a<0) {putchar(‘-‘); a=-a;} if(a>=10) Out(a/10); putchar(a%10+‘0‘); } const int N=1005; //Code begin... int dp[N][N], len1, len2; char s1[N], s2[N]; int dfs(int x, int y) { if (~dp[x][y]) return dp[x][y]; if (x==len1&&y==len2) return 0; if (x==len1) return len2-y; if (y==len2) return len1-x; int ans=dfs(x+1,y)+1; ans=min(ans,dfs(x+1,y+1)+(s1[x]==s2[y]?0:1)); ans=min(ans,dfs(x,y+1)+1); return dp[x][y]=ans; } int main () { while (~scanf("%d%s%d%s",&len1,s1,&len2,s2)) { mem(dp,-1); printf("%d\n",dfs(0,0)); } return 0; }
标签：can 匹配 cto div tac 1.0 inf putchar return
原文地址：http://www.cnblogs.com/lishiyao/p/6426241.html
