HDU 5045 Contest expect + pressure dp 2014 ACM/ICPC Asia Regional Shanghai Online, hduicpc

HDU 5045 Contest expect + pressure dp 2014 ACM/ICPC Asia Regional Shanghai Online, hduicpc

Question:

Given n people m questions

The n * m matrix below indicates the probability of each person solving each question.

We can get a set of m values {1, 2, 3, 1, 2}

Indicates who solved each question.

There are many sets, and there is an expectation for this set, and the set with the largest expectation (what is the expectation)

One restriction: This is invalid for the set {, 1} (that is, the question from [1, n] must be 1-n, and then [n + 1, 2n] question is also an arrangement)

Then press dp

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <vector>using namespace std;typedef long long ll;const int M= 1000 +2;const int N= 10+2;const int S= (1 << 10) + 10;double d[M][S], p[N][M];int main() {    int cas, mx, n, m, to;    scanf("%d", &cas);    for(int T= 1; T<= cas; ++T) {        scanf("%d%d", &n, &m);        for (int i = 0; i < n; ++i)            for (int j = 0; j < m; ++j)                scanf("%lf", &p[i][j]);        mx = (1 << n) - 1;        memset(d, 0, sizeof d);        d[0][0] = 0;        for (int i = 0; i < m; ++i) {            int s = (1 << (i % n)) - 1;            while (s < mx) {                int x = s & -s;                int y = s + x;                for (int k = 0; k < n; ++k)                    if (!(s >> k & 1)) {                        to = s | (1 << k);                        if (to == mx)                            to = 0;                        d[i + 1][to] = max(d[i + 1][to], d[i][s] * (1.0 - p[k][i]) + (d[i][s] + 1) * p[k][i]);                    }                if (s == 0) break;                s = ((s & ~y) / x >> 1) | y;            }        }        double ans = 0;        for (int i = 0; i < mx; ++i)            ans = max(ans, d[m][i]);        printf("Case #%d: %.5f\n", T, ans);    }    return 0;}

2012 how many teams are ACM/ICPC Asia Regional Chengdu Online?

130 teams
 
The 2009 ACM-ICPC Asia Ningbo Regional Online Contest-D

Haha ...... I am also doing this. Let's talk about it, that is, using a diamond (defined as a diamond in the distance from Manhattan), a square with 45 degrees of rotation, a nested one, and finding the center of the circle.