设为首页 加入收藏

TOP

洛谷P2973 [USACO10HOL]赶小猪(高斯消元 期望)
2019-01-05 20:09:03 】 浏览:37
Tags:洛谷 P2973 USACO10HOL 高斯 期望

题意

题目链接

Sol

\(f[i]\)表示炸弹到达\(i\)这个点的概率,转移的时候考虑从哪个点转移而来

\(f[i] = \sum_{\frac{f(j) * (1 - \frac{p}{q})}{deg(j)}}\)

\(f[1]\)需要+1(炸弹一开始在1)

// luogu-judger-enable-o2
#include<bits/stdc++.h> 
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long 
#define LL long long 
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 3001, mod = 1e9 + 7, INF = 1e9 + 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename A> inline void debug(A a){cout << a << '\n';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M, vis[MAXN][MAXN], deg[MAXN];
double P, Q, a[MAXN][MAXN];
void Gauss() {
    for(int i = 1; i <= N; i++) {
        int mx = i;
        for(int j = i + 1; j <= N; j++) if(a[j][i] > a[mx][i]) mx = j;
        if(mx != i) swap(a[i], a[mx]);
        for(int j = i + 1; j <= N + 1; j++) a[i][j] /= a[i][i]; a[i][i] = 1;
        for(int j = 1; j <= N; j++) {
            if(i == j) continue;
            double p = a[j][i] / a[i][i];
            for(int k = 1; k <= N + 1; k++) {
                a[j][k] -= a[i][k] * p;
            }
        }
    }
}
signed main() {
    N = read(); M = read(); P = read(); Q = read();
    for(int i = 1; i <= M; i++) {
        int x = read(), y = read();
        vis[x][y] = vis[y][x] = 1;
        deg[x]++; deg[y]++;
    }
    for(int i = 1; i <= N; i++) {
        a[i][i] = 1;
        for(int j = 1; j <= N; j++) 
            if(vis[i][j]) 
                a[i][j] = -(1.0 - P / Q) / deg[j];      
    }
    a[1][N + 1] = 1;
    Gauss();
    for(int i = 1; i <= N; i++) printf("%.9lf\n", a[i][N + 1] * (P / Q));
    return 0;
}
/*

5 4
1 2 2 1 3
1 3

*/

编程开发网
】【打印繁体】【投稿】【收藏】 【推荐】【举报】【评论】 【关闭】 【返回顶部
上一篇写高并发程序时慎用strncpy和spri.. 下一篇C++11程序设计要点总结-模板机制..

评论

帐  号: 密码: (新用户注册)
验 证 码:
表  情:
内  容:

array(4) { ["type"]=> int(8) ["message"]=> string(24) "Undefined variable: jobs" ["file"]=> string(32) "/mnt/wp/cppentry/do/bencandy.php" ["line"]=> int(214) }