洛谷P2973 [USACO10HOL]赶小猪(高斯消元 期望)
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2022-11-10 22:19:25
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) cpp // luogu judger enabl ......
题意
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)
// 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 */
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