区间DP专辑 不断更新中

POJ 1651：记N个卡片为num[1]到num[N]，容易发现取到最后总会剩下头尾两张卡片，就是说最后一次取的时候只剩下三张，只能取中间那张，所以将最后取走的那张卡片的位置当做区间的分割点，将问题分为左右两个区间的两个子问题，并且都满足最优子结构性质。记第i张卡片到第j张卡片的最优解为DP(i, j)，那么可以得到状态转移方程：

DP(i, j) = 0, 如果i = j - 1;

DP(i, j) =  min{DP(i, k) + DP(k, j) + num[i] * num[k] * num[j] } , 其中i < k < j， 且j - i >= 3.

注意这题让求最小值，一开始把dp初始化成一个大数。

#include <stdio.h>#include <string.h>#define inf 0x3f3f3f3f  #define MAX 105int min (int a, int b){return a > b ? b : a;}int main (){int dp[MAX][MAX], num[MAX];int i, j ,k, sublen;int cases;scanf("%d", &cases);for(i = 0; i < cases; i++){scanf("%d", &num[i]);}for(i = 0; i < cases; i++){for(j = 0; j < cases; j++){if(i + 1 == j){dp[i][j] = 0;}else{dp[i][j] = inf;}}}for(sublen = 2; sublen < cases; sublen++){for(i = 0, k = sublen; k < cases; i++, k++){for(j = i + 1; j < k; j++){dp[i][k] = min(dp[i][k], dp[i][j] + dp[j][k] + num[i] * num[j] * num[k]);}}}printf("%d\n", dp[0][cases-1]);return 0;}

POJ 2955：最大括号匹配，先放代码，明天写题解。

#include <stdio.h>#include <string.h>#define MAX 110int max (int a, int b){return a > b ? a : b;}int main (){char seq[MAX];int dp[MAX][MAX];int len;int i, j, k, sublen;while (scanf("%s", seq)){if ('e' == seq[0]){break;}memset(dp, 0, sizeof(dp));len = strlen(seq);for (i = 0; i < len - 1; i++){if (('(' == seq[i] && ')' == seq[i+1]) || ('[' == seq[i] && ']' == seq[i+1])){dp[i][i+1] = 2;}}for (sublen = 2; sublen < len; sublen++){for (i = 0, j = i + sublen; j < len; i++, j++){/*先假设seq[i]和后面没有匹配成功  则dp[i][j] = dp[i+1][j]*/dp[i][j] = dp[i+1][j];for (k = i + 1; k <= j; k++){/*如果seq[i]和后面匹配成功  更新dp[i][j]的值*/if (('(' == seq[i] && ')' == seq[k])  //|| ('[' == seq[i] && ']' == seq[k])){dp[i][j] = max(dp[i][j], dp[i+1][k-1] + dp[k+1][j] + 2);}}}}printf("%d\n", dp[0][len-1]);}return 0;}

HihoCoder 1110：正则表达式  解题报告