BZOJ4008 [HNOI2015]亚瑟王 【概率dp】

题目链接

BZOJ4008

题解

要求所有牌造成伤害的期望，就是求每一张牌发动的概率\(g[i]\)

我们发现一张牌能否发动，还与其前面的牌是否发动有关

那我们设\(f[i][j]\)表示前\(i\)张在\(r\)轮游戏中总共发动了\(j\)张的概率

那么

\[g[i] = \sum\limits_{j = 0}^{min\{i - 1,r\}} f[i - 1][j](1 - (1 - p[i])^{r - j})
\]

因为前\(i - 1\)张发动了\(j\)个，一定占用了\(j\)轮，剩余\(r - j\)轮中\(i\)必须发动

考虑如何求\(f[i][j]\)

同样是考虑\(i\)发动不发动

如果发动

\[f[i][j] += f[i - 1][j - 1](1 - (1 - p[i])^{r - j + 1})
\]

如果不发动

\[f[i][j] += f[i - 1][j](1 - p[i])^{r - j}
\]

这样这题就做完了

时间复杂度\(O(Tnr)\)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 100005,maxm = 100005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
double f[222][135],g[222],p[222],d[222],pw[222][135];
int n,r;
int main(){
	int T = read();
	while (T--){
		n = read(); r = read();
		REP(i,n){
			scanf("%lf",&p[i]);
			d[i] = read();
			pw[i][0] = 1;
			for (int j = 1; j <= r; j++)
				pw[i][j] = pw[i][j - 1] * (1 - p[i]);
		}
		f[0][0] = 1;
		for (int i = 1; i <= n; i++)
			for (int j = 0; j <= min(i,r); j++){
				f[i][j] = 0;
				if (j) f[i][j] += f[i - 1][j - 1] * (1 - pw[i][r - j + 1]);
				if (i > j) f[i][j] += f[i - 1][j] * pw[i][r - j];
		}
		double ans = 0;
		for (int i = 1; i <= n; i++){
			g[i] = 0;
			for (int j = 0; j <= min(i - 1,r); j++)
				g[i] += f[i - 1][j] * (1 - pw[i][r  - j]);
			ans += g[i] * d[i];
		}
		printf("%.10lf\n",ans);
	}
	return 0;
}