Peter's Hobby
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题目链接
- 题意:
题意比较麻烦。。。n天,给出每天的叶子的一种状态(Dry , Dryish , Damp and Soggy),最有可能出现的天气序列(Sunny, Cloudy and Rainy)
最开始,第一天处于每种状态有一个预定义的概率。每一天,根据当前的叶子状态,处于每种天气状况有一个给定的概率;题目中还给出任意两种天气状态的转移概率,即前一天处于某种天气时今天处于某种天气的概率。 - 分析:
就是简单的DP,每天处于每个点有一个固定的状态,前一天的每个状态的转移概率也是给定的,起始点也给定,所以就需要处理一下这些即可。除此之外,题目中需要输出“路径”,多记录一下即可
const int maxn = 100010;double dp[maxn][3];int ipt[maxn], p[maxn][3];char s[10];double d1[][4] ={ {0.6, 0.2, 0.15, 0.05}, {0.25, 0.3, 0.2, 0.25}, {0.05, 0.1, 0.35, 0.5}};double d2[][3] ={ {0.5, 0.375, 0.125}, {0.25, 0.125, 0.625}, {0.25, 0.375, 0.375}};map<string, int> mp;char to[][10] = {"Sunny", "Cloudy", "Rainy"};int main(){ mp["Dry"] = mp["Sunny"] = 0; mp["Dryish"] = mp["Cloudy"] = 1; mp["Damp"] = mp["Rainy"] = 2; mp["Soggy"] = 3; int T, n; RI(T); FE(kase, 1, T) { CLR(p, -1); RI(n); REP(i, n) { RS(s); ipt[i] = mp[s]; } dp[0][0] = log(0.63) + log(d1[0][ipt[0]]); dp[0][1] = log(0.17) + log(d1[1][ipt[0]]); dp[0][2] = log(0.2) + log(d1[2][ipt[0]]); FF(i, 1, n) { REP(j, 3) { dp[i][j] = -1e10; REP(k, 3) { double pre = dp[i - 1][k] + log(d2[k][j]) + log(d1[j][ipt[i]]); if (pre > dp[i][j]) { dp[i][j] = pre; p[i][j] = k; } } } } double Max = -1e10; int c = 0; REP(j, 3) if (dp[n - 1][j] > Max) { Max = dp[n - 1][j]; c = j; } stack<int> sk; int r = n - 1; while (r >= 0) { sk.push(c); c = p[r--][c]; } printf("Case #%d:\n", kase); while (!sk.empty()) { printf("%s\n", to[sk.top()]); sk.pop(); } } return 0;}
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