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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;
}
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原文地址：http://www.cnblogs.com/Running-Time/p/4907405.html
