2013长沙现场赛J - Josephina and RPG
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题意:给你各个队伍对战的胜率表,然后给你要按顺序挑战的队伍,问全胜的最大胜率是多少。这道题的难点在于,获胜的时候可以选择更换队伍,也可以不更换队伍,一开始你可以选择任意队伍。
思路:这道题是一道简单的dp题目,每次只有C(m,3)种状态,每种状态表示当前队伍是哪个,我们用now数组来表示,然后每次有两种选择,更换队伍或不更换队伍。不更换队伍很明显就是now[j] *= t[j][x[i]];更换队伍的是用前一个状态中的最大值去乘t[x[i - 1]][x[i]],能理解的话基本上这道题就完了。下面给代码。
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm>#include<string>#include<queue>#include<cmath>#include<set>using namespace std;#define maxn 10005#define xx 1000000001double t[1005][1005], now[1005];int main() {int m;int a[15];a[0] = 1;a[1] = 1;for (int i = 2; i <= 10; i++) {a[i] = a[i - 1] * i;}while (scanf("%d", &m) != EOF) {int r = a[m] / a[m - 3] / 6;for (int i = 0; i < r; i++) {for (int j = 0; j < r; j++) {scanf("%lf", &t[i][j]);}}int n;int x[maxn];for (int i = 0; i < r; i++) {now[i] = 1;}scanf("%d", &n);scanf("%d", &x[0]);for (int j = 0; j < r; j++) {now[j] *= t[j][x[0]];}for (int i = 1; i < n; i++) {scanf("%d", &x[i]);double maxnum = 0;for (int j = 0; j < r; j++) {maxnum = max(now[j], maxnum);if (j != x[i - 1]) {now[j] *= t[j][x[i]];}}now[x[i - 1]] = maxnum*t[x[i - 1]][x[i]];}double maxnum = 0;for (int i = 0; i < r; i++) {maxnum = max(maxnum, now[i]);}printf("%.6lf\n", maxnum);}}
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