kuangbin专题二十二:区间DP

 

POJ2955 Brackets

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 105;
char s[maxn];
int dp[maxn][maxn];

int main(){
    while(~scanf("%s", s) && s[0] != 'e') {
        memset(dp, 0, sizeof(dp));

        int n = strlen(s);
        for(int len = 1; len <= n; len++) {
            for(int i =  0; i < n; i++) {
                if(i + len - 1 < n) {
                    int j = i + len - 1;
                    if((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']'))
                        dp[i][j] = dp[i + 1][j - 1] + 2;
                    for(int k = i; k < j; k++)
                        dp[i][j] = max(dp[i][j], dp[i][k] + dp[k + 1][j]);
                }
            }
        }

        printf("%d\n", dp[0][n - 1]);
    }
    return 0;
}
View Code

 

POJ1651 Multiplication Puzzle

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 105;
int ary[maxn];
int dp[maxn][maxn];

int main(){
    int n;
    cin >> n;
    for(int i = 1; i <= n; i++) cin >> ary[i];


    for(int len = 3; len <= n; len++) {
        for(int i = 1; i <= n - len + 1; i++) {
            int j = i + len - 1;
            dp[i][j] = 0x3f3f3f3f;
            for(int k = i + 1; k < j; k++){
                dp[i][j] = min(dp[i][j], dp[i][k] + ary[i] * ary[k] * ary[j] + dp[k][j]);
            }
        }
    }

    cout << dp[1][n] << endl;

    return 0;
}
View Code

 

posted @ 2021-07-24 21:42  Nanachi  阅读(34)  评论(0)    收藏  举报