题目链接:http://poj.org/problem?id=3280
思路: dp[i][j] :=第i个字符到第j个字符之间形成回文串的最小费用。
dp[i][j]=min(dp[i+1][j]+cost[s[i-1]-'a'],dp[i][j-1]+cost[s[j-1]-'a']);
if(s[i-1]==s[j-1]) dp[i][j]=min(dp[i+1][j-1],dp[i][j]);
注意循环顺序,我觉得这题就是这里是tricky:
#include<iostream> #include<cstring> #include<cstdlib> #include<algorithm> using namespace std; const int MAX_M=2005; int dp[MAX_M][MAX_M],cost[26]; int main() { int N,M; while(cin>>N>>M) { string s; cin>>s; for(int i=0;i<N;i++){ char ch; int ca,cb; cin>>ch>>ca>>cb; cost[ch-'a']=min(ca,cb); } memset(dp,0,sizeof(dp)); for(int j=1;j<=M;j++){ for(int i=j-1;i>=1;i--){ dp[i][j]=min(dp[i+1][j]+cost[s[i-1]-'a'],dp[i][j-1]+cost[s[j-1]-'a']); if(s[i-1]==s[j-1]) dp[i][j]=min(dp[i+1][j-1],dp[i][j]); } } cout<<dp[1][M]<<endl; } return 0; }