前往小程序,Get更优阅读体验!
立即前往
首页
学习
活动
专区
工具
TVP
发布
社区首页 >专栏 >洛谷P2973 [USACO10HOL]赶小猪(高斯消元 期望)

洛谷P2973 [USACO10HOL]赶小猪(高斯消元 期望)

作者头像
attack
发布2019-01-30 16:44:07
3780
发布2019-01-30 16:44:07
举报
文章被收录于专栏:数据结构与算法

题意

题目链接

Sol

设\(f[i]\)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)

代码语言:javascript
复制
// 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

*/
本文参与 腾讯云自媒体同步曝光计划,分享自作者个人站点/博客。
原始发表:2019-01-05 ,如有侵权请联系 cloudcommunity@tencent.com 删除

本文分享自 作者个人站点/博客 前往查看

如有侵权,请联系 cloudcommunity@tencent.com 删除。

本文参与 腾讯云自媒体同步曝光计划  ,欢迎热爱写作的你一起参与!

评论
登录后参与评论
0 条评论
热度
最新
推荐阅读
目录
  • 题意
  • Sol
领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档