DP ZOJ 3735 Josephina and RPG

题目传送门

题意：告诉你C(m,3)个队伍相互之间的胜率，然后要你依次对战n个AI队伍，首先任选一种队伍，然后战胜一个AI后可以选择替换成AI的队伍，也可以不换，问你最后最大的胜率是多少。

分析：dp[i][j][0/1] 表示第i个AI，用j的id去攻打，此j可以是上一个状态交换AI的id而来也可以不是，状态转移方程：

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]]);

效率并不是很高。。。

/************************************************
* Author        :Running_Time
* Created Time  :2015/10/24 星期六 13:32:44
* File 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 | 1
typedef long long ll;
const int N = 1e4 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
double dp[N][122][2];
double 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 == 10)   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;
}