DP 之 poj 3280

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//  [4/4/2014 Sjm]/*状态： dp[i][j] := 从 i 位置到 j 位置构成回文串，所需要的最小开销决策:1) 若 str[i] == str[j], 则此时可获得 dp[i][j] 的一种情况，即 dp[i][j] = dp[i+1][j-1] 2）若 str[i] ！= str[j] 或 str[i] == str[j] 皆可能有以下操作：1、可删除 str[i]，此时可获得 dp[i][j] 的一种情况，即 dp[i][j] = dp[i+1][j] + mymap[str[i]].del_Value2、可在 j 的后面添加 str[i], 此时可获得 dp[i][j] 的一种情况，即 dp[i][j] = dp[i+1][j] + mymap[str[i]].add_Value与 1, 2 同，亦可对 str[j] 进行处理，故不再详述*/
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <climits>#include <map>using namespace std;const int MAX_N = 2000;int dp[MAX_N][MAX_N], N, mylen;char str[MAX_N];struct node{int add_Value, del_Value;};map<char, node> mymap;int Get_Min(int i, int j){int temp = INT_MAX;temp = min(temp, dp[i + 1][j] + min(mymap[str[i]].del_Value, mymap[str[i]].add_Value));temp = min(temp, dp[i][j - 1] + min(mymap[str[j]].del_Value, mymap[str[j]].add_Value));return temp;}int Solve(){memset(dp, 0, sizeof(dp));for (int t = 1; t < mylen; t++){for (int i = 0; i < (mylen - t); i++) {int j = i + t;dp[i][j] = INT_MAX;if (str[i] == str[j])dp[i][j] = min(dp[i][j], dp[i + 1][j - 1]);dp[i][j] = min(dp[i][j], Get_Min(i, j));}}return dp[0][mylen - 1];}int main(){//freopen("input.txt", "r", stdin);//freopen("output.txt", "w", stdout);scanf("%d %d", &N, &mylen);scanf("%s", str);for (int i = 0; i < N; i++) {char c;int aValue, dValue;cin >> c >> aValue >> dValue;mymap[c].add_Value = aValue;mymap[c].del_Value = dValue;}printf("%d\n", Solve());return 0;}