LA6878

区间dp

dp[i][j]存i->j区间的所有取值

然后枚举分割点，枚举两个存的值，分别运算存储。

看见这种不确定分割顺序，两个区间合并的情况，就要用区间dp。

#include<bits/stdc++.h>
using namespace std;
const int N = 60;
int n, kase;
int vis[N * 110];
char s[N];
vector<int> dp[N][N];
int main()
{
    while(scanf("%s", s + 1))
    {
        if(s[1] == '0') break;
        n = strlen(s + 1);
        for(int i = 1; i <= n; ++i)
            for(int j = i; j <= n; ++j)
                dp[i][j].clear();
        for(int i = 1; i <= n; ++i)
        {
            if(s[i] == 'I') dp[i][i].push_back(1);
            if(s[i] == 'V') dp[i][i].push_back(5);
            if(s[i] == 'X') dp[i][i].push_back(10);
            if(s[i] == 'L') dp[i][i].push_back(50);
            if(s[i] == 'C') dp[i][i].push_back(100);
        }
        for(int len = 1; len <= n; ++len)
            for(int i = 1; i <= n - len + 1; ++i)
            {
                for(int j = i; j < i + len; ++j)
                    for(int x = 0; x < dp[i][j].size(); ++x)
                        for(int y = 0; y < dp[j + 1][i + len - 1].size(); ++y)
                            if(dp[i][j][x] < dp[j + 1][i + len - 1][y] && !vis[dp[j + 1][i + len - 1][y] - dp[i][j][x]])
                            {
                                dp[i][i + len - 1].push_back(dp[j + 1][i + len - 1][y] - dp[i][j][x]);
                                vis[dp[j + 1][i + len - 1][y] - dp[i][j][x]] = 1;
                             }
                            else if(dp[i][j][x] >= dp[j + 1][i + len - 1][y] && !vis[dp[j + 1][i + len - 1][y] + dp[i][j][x]])
                            {
                                vis[dp[j + 1][i + len - 1][y] + dp[i][j][x]] = 1;
                                dp[i][i + len - 1].push_back(dp[j + 1][i + len - 1][y] + dp[i][j][x]);
                            }
                    for(int x = 0; x < dp[i][i + len - 1].size(); ++x)
                        vis[dp[i][i + len - 1][x]] = 0;
            }
        sort(dp[1][n].begin(), dp[1][n].end());
        printf("Case %d:", ++kase);
        for(int i = 0; i < dp[1][n].size(); ++i)
            printf(" %d", dp[1][n][i]);
        puts("");
    }
    return 0;
}