DP ZOJ 3735 Josephina and RPG

Topic Portal

Test instructions: Tell you C (m,3) a team between each other's winning percentage, and then you want to play against N AI team, the first choice of a team, and then overcome an AI can choose to replace the AI team, you can also do not change, ask you the last biggest winning percentage is how much.

Analysis: DP[I][J][0/1] represents the first AI, with the ID of J to attack, this J can be the ID of the previous state exchange AI or not, the state transition equation:

Dp[i][j][0] = max (dp[i-1][j][0], dp[i-1][j][1]) * P[j][a[i]];
if (i > 1)
DP[I][A[I-1]][1] = max (dp[i][a[i-1]][1], Max (dp[i-1][j][0], dp[i-1][j][1]) * P[a[i-1]][a[i]]);

Efficiency is not very high ...

/************************************************* author:running_time* Created time:2015/10/24 Saturday 13:32:44* Fi Le name:j.cpp ************************************************/#include <cstdio> #include <algorithm># Include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string > #include <vector> #include <queue> #include <deque> #include <stack> #include <list># Include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime>using namespace std; #define Lson L, Mid, RT << 1#define Rson mid + 1, R, RT << 1 | 1typedef long ll;const int N = 1e4 + 10;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const Double EPS = 1e-8;do    Uble dp[n][122][2];d ouble p[122][122];int a[n];int Main (void) {int m, r, N;        while (scanf ("%d", &m) = = 1) {if (M = = 3) R = 1;        else if (m = = 4) R = 4; else if (m = = 5)    R = 10;        else if (m = = 6) R = 20;        else if (m = = 7) R = 35;        else if (m = = 8) R = 56;        else if (m = = 9) R = 84;        else if (m = = Ten) R = 120;            for (int. I=1; i<=r; ++i) {for (int j=1; j<=r; ++j) {scanf ("%lf", &p[i][j]);        }} scanf ("%d", &n);    for (int i=1; i<=n; ++i) {scanf ("%d", &a[i]);        a[i]++;        } memset (DP, 0, sizeof (DP));        for (int i=1; i<=r; ++i) dp[0][i][0] = dp[0][i][1] = 1; for (int i=1, i<=n; ++i) {for (int j=1; j<=r; ++j) {dp[i][j][0] = max (dp[i-1][j][                0], dp[i-1][j][1]) * P[j][a[i]; if (i > 1) dp[i][a[i-1]][1] = max (dp[i][a[i-1]][1], Max (dp[i-1][j][0], dp[i-1][j][1]) * p[a[i-1]][            A[i]]);        }} double ans = 0; for (int i=1; i<=r; ++i) ans = max (ans, Max (dp[n][i][0], dp[n][i)[1]));    printf ("%.6f\n", ans); } return 0;}

　　

DP ZOJ 3735 Josephina and RPG