UVA 10417 - Gift Exchanging(概率 DFS）

最新推荐文章于 2022-05-02 11:18:05 发布

原创最新推荐文章于 2022-05-02 11:18:05 发布 · 1k 阅读

0 ·

CC 4.0 BY-SA版权

在一个圣诞派对上，每位参与者带来一份礼物，参与者需从中挑选一份带回家。作为第一个选择的人，如何通过了解每位朋友携带不同礼物箱的概率来最大化获得心仪礼物的机会。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem G
Gift Exchanging
Input: standard input
Output: standard output
Time Limit: 15 seconds
Memory Limit: 64 MB

You're attending a Xmas party, every one there (excepty you, I don't know why...) has brought a beautiful little gift in a box. After all the gifts are collected and laid on the table, every one can choose one to take home. You are the first one to choose, so there is a big chance for you to get the gift that is brought by your best friend :-)

Unfortunately, you cannot see inside the gift box, but luckily, there are only 5 possible kinds of boxes allowed in the party. We useA,B,C,D,E to represent these five kinds. You know well about all the guests there. i.e you know the probability that each person brings each kind of box. Which box should you take home, and what is the probability that you get the one you want?

Input

The first line of the input is a single integer t(1<=t<=10), indicating the number of test cases. Each case begins with a line containing a single integer n(1<=n<=12), indicating the number of guests. The second line of the case contains 5 integers indicating the number of gifts of each kind. The sum of these 5 numbers always equals to n. The following n lines each contain 5 non-negative floating numbers(with one digit after decimal point) indicating the probabilities for him to take box A,B,C,D,E. The sum of 5 numbers always equals to 1.0. The first guest is your best friend.

Output

For each test case, print a single character indicating the box you should choose, then a single space, then the probability to 3 decimal places. If there are more than one best choices, print the smallest box number.

Sample Input
2
2
1 0 0 1 0
0.5 0.0 0.0 0.5 0.0
0.4 0.0 0.5 0.1 0.0
8
1 3 2 1 1
0.1 0.2 0.3 0.2 0.2
0.0 0.8 0.2 0.0 0.0
0.3 0.0 0.1 0.6 0.0
0.0 0.9 0.1 0.0 0.0
0.2 0.2 0.0 0.3 0.3
0.2 0.1 0.6 0.1 0.0
0.7 0.2 0.1 0.0 0.0
0.1 0.0 0.0 0.2 0.7

Sample Output
4 0.800
3 0.296

题意：n个小伙伴每人准备了1个礼物，第一行为组合出的情况res，后面n行为小伙伴送每种礼物的概率。要求出选那种礼物，拿到第一个小伙伴给的概率最高

思路:概率，以组合出res情况的概率做分母，然后枚举第一个小伙伴选每种礼物包含在其中的概率。去进行深搜即可

代码:

#include <stdio.h>
#include <string.h>

int t, n, res[5];
double mu, zi[5], map[12][5];

void dfs(int num, double p, int j) {
    if (num == n) {
	zi[j] += p;
    }
    for (int i = 0; i < 5; i ++) {
	if (res[i] == 0) continue;
	res[i] --;
	dfs(num + 1, p * map[num][i], j);
	res[i] ++;
    }
}

int main() {
    scanf("%d", &t);
    while (t --) {
	memset(zi, 0, sizeof(zi));
	mu = 0;
	scanf("%d", &n);
	for (int i = 0; i < 5; i ++)
	    scanf("%d", &res[i]);
	for (int i = 0; i < n; i ++)
	    for (int j = 0; j < 5; j ++)
		scanf("%lf", &map[i][j]); 
	for (int i = 0; i < 5; i ++) {
	    if (res[i] == 0) continue;
	    res[i] --;
	    dfs(1, map[0][i], i);
	    res[i] ++;
	}
	for (int i = 0; i < 5; i ++)
	    mu += zi[i];
	double ans = 0; int v = 0;
	for (int i = 0; i < 5; i ++) {
	    if (res[i] == 0) continue;
	    if (ans < zi[i] / (mu * res[i])) {
		ans = zi[i] / (mu * res[i]);
		v = i + 1;
	    }
	}
	printf("%d %.3lf\n", v, ans);
	
    }
    return 0;
}