POJ 1179

哎 悲剧啊 竟然搞了4个小时！

 

枚举断边  再dp

 

将存数组和符号的数组扩大一倍能使思维更简单

 

特别注意负负得正的情况  所以max min都要记下来

 

  #include<iostream>using namespace std;#define MAXN 101struct Node {long mmax;long mmin;} f[MAXN][MAXN];int main() {long n, num[101], l, i, k, j, len, maxx[101], want=-65535;char op[101];cin >> n;for (i = 1;i <= n;i++) {cin >> op[i] >> num[i];num[i + n] = num[i];op[i + n] = op[i];}for (l = 1; l <= n; l++) {for (k = 1; k <= 100; k++) {for (j = 1; j <= 100; j++) {f[k][j].mmax = -65535;f[k][j].mmin = 65535;}}for (k = l; k <= l + n - 1; k++) {f[k][k].mmax = f[k][k].mmin = num[k];}for (len = 2; len <= n; len++)for (i = l; i <= l + n - 1; i++) {j = i + len - 1;if (j > l + n - 1)break;for (k = i + 1; k <= j; k++) {if (op[k] == 't') {f[i][j].mmax = max(f[i][j].mmax,f[i][k - 1].mmax + f[k][j].mmax);f[i][j].mmin = min(f[i][j].mmin,f[i][k - 1].mmin + f[k][j].mmin);} else {f[i][j].mmax = max(f[i][j].mmax, max(f[i][k - 1].mmax * f[k][j].mmax,max(f[i][k - 1].mmin * f[k][j].mmax,max(f[i][k - 1].mmax* f[k][j].mmin, f[i][k- 1].mmin* f[k][j].mmin))));f[i][j].mmin = min(f[i][j].mmin, min(f[i][k - 1].mmax * f[k][j].mmax,min(f[i][k - 1].mmin * f[k][j].mmax,min(f[i][k - 1].mmax* f[k][j].mmin, f[i][k- 1].mmin* f[k][j].mmin))));}}}maxx[l] = f[l][l + n - 1].mmax;want = max(want, maxx[l]);}cout << want << endl;for (l = 1; l <= n; l++)if (maxx[l] == want)cout << l <<" ";cout << endl;return 0;}