演算法導論-15-3-編輯距離
阿新 • • 發佈:2019-02-20
題目:
思考:
(1)初始化:
s[0,0]=0
s[i,0] = i * cost(delete)
s[0,j] = j * cost[insert]
(2)遞推
(3)最後的操作kill
c[i][j] = MIN(c[m,n], MIN(c[i,n]+cost(kill))), 其中0<=i<m
程式碼:
#include <iostream> using namespace std; //6種操作的代價,依次是copy,replace,delete,insert,twiddle,kill int cost[6] = {1,1,1,1,1,1}; int s[15][15] = {0}; //字元從0開始計數,s[][]從1開始 int DP(char *x, char *z, int lx, int lz) { int i = 0, j = 0,temp; //初始化s[0][0]=0 //s[i,0] = i * cost(delete) for(i = 1; i <= lx; i++) s[i][0] = i * cost[2]; //s[0,j] = j * cost[insert] for(j = 1; j <= lz; j++) s[0][j] = i * cost[3]; //DP for(i = 0; i < lx; i++) { for(j = 0; j < lz; j++) { s[i+1][j+1] = 0x7fffffff; //copy if(x[i] == z[j]) { temp = s[i][j] + cost[0]; if(temp < s[i+1][j+1]) s[i+1][j+1] = temp; } //replace if(x[i] != z[j]) { temp = s[i][j] + cost[1]; if(temp < s[i+1][j+1]) s[i+1][j+1] = temp; } //delete temp = s[i][j+1] + cost[2]; if(temp < s[i+1][j+1]) s[i+1][j+1] = temp; //insert temp = s[i+1][j] + cost[3]; if(temp < s[i+1][j+1]) s[i+1][j+1] = temp; //twidle if(i && j && x[i] == z[j-1] && x[i-1] == z[j]) { temp = s[i-1][j-1] + cost[4]; if(temp < s[i+1][j+1]) s[i+1][j+1] = temp; } } } //kill int ret = s[lx][lz]; for(i = 1; i < lx; i++) if(s[i][lz] + cost[5] < ret) ret = s[i][lz] + cost[5]; return ret; } void Print(int lx, int lz) { int i, j; for(i = 1; i <= lx; i++) { for(j = 1; j <= lz; j++) cout<<s[i][j]<<' '; cout<<endl; } } int main() { char a[] = "algorithm"; char b[] = "altruistic"; int la = strlen(a), lb = strlen(b); cout<<DP(a, b, la, lb)<<endl; Print(la, lb); return 0; }