标签：path ace back cstring ctime pac struct print min
给定一张 $n$ 个点 $m$ 条边的无向图.
每条边有两个信息 $x, y$ , 这条边的边权是 $ax + (1 - a)y$ .
给定源点 $S$ 和汇点 $T$ , 求当 $a \in [0, 1]$ 时, $E(minpath(S, T))$ .
$n \le 200, m \le 400$ .
不论是期望还是概率, 都是在某种量化意义下, 将试验次数取无穷大.
我们均匀取点, 进行 SPFA , 然后取平均值.
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cctype> 5 #include <cmath> 6 #include <ctime> 7 #include <vector> 8 using namespace std; 9 #define F(i, a, b) for (register int i = (a); i <= (b); i++) 10 #define fore(it, x) for (register vector<Edge>::iterator it = g[x].begin(); it != g[x].end(); it++) 11 #define db double 12 inline int rd(void) { 13 int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -1; 14 int x = 0; for (; isdigit(c); c = getchar()) x = x*10+c-‘0‘; return x*f; 15 } 16 17 const db EPS = 1e-7; 18 inline int sign(db x) { return fabs(x) < EPS ? 0 : x < 0 ? -1 : 1; } 19 inline int cmp(db x, db y) { return sign(x - y); } 20 21 const int N = 205; 22 const db INF = 1e20; 23 24 int n, m, S, T; 25 struct Edge { 26 int v; db x, y; 27 inline db d(db a) { return a * x + (1 - a) * y; } 28 }; 29 vector<Edge> g[N]; 30 31 inline void Init(int s, int t, db x, db y) { 32 g[s].push_back((Edge){t, x, y}); 33 g[t].push_back((Edge){s, x, y}); 34 } 35 36 int nT; db res; 37 int q[N], qh, qt; bool inq[N]; db dis[N]; 38 39 db SPFA(db a) { 40 memset(q, 0, sizeof q), qh = qt = 0, memset(inq, false, sizeof inq), fill(dis + 1, dis + n + 1, INF); 41 q[++qt] = S, inq[S] = true, dis[S] = 0; 42 while (qh != qt) { 43 int x = q[qh = qh % n + 1]; 44 fore(it, x) 45 if (cmp(dis[x] + it->d(a), dis[it->v]) < 0) { 46 dis[it->v] = dis[x] + it->d(a); 47 if (!inq[it->v]) 48 q[qt = qt % n + 1] = it->v, inq[it->v] = true; 49 } 50 inq[x] = false; 51 } 52 return dis[T]; 53 } 54 55 int main(void) { 56 #ifndef ONLINE_JUDGE 57 // freopen("C.in", "r", stdin); 58 #endif 59 60 srand(time(0)); 61 62 n = rd(), m = rd(), S = rd(), T = rd(); 63 F(i, 1, m) { 64 int s = rd(), t = rd(); 65 db x, y; scanf("%lf %lf", &x, &y); 66 Init(s, t, x, y); 67 } 68 69 nT = 100000000 / (5 * n); 70 db V = 1.0 / (nT-1), a = 0; // seperated equally 71 for (int t = 1; t <= nT; t++) { 72 res += (1.0 / nT) * SPFA(a); 73 a += V; 74 } 75 printf("%0.10lf\n", res); 76 77 return 0; 78 }
这道题要求连续性随机变量的数学期望.
设当 $a = x$ 时, 最短路的长度是 $f_a$ .
那么我们要求 $\frac{\int_0 ^ 1 f_a}{1 - 0}$ .
Simpson积分套最短路.
1 #include <cstdio> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cctype> 5 #include <cmath> 6 #include <vector> 7 #include <map> 8 using namespace std; 9 #define F(i, a, b) for (register int i = (a); i <= (b); i++) 10 #define fore(it, x) for (register vector<Edge>::iterator it = g[x].begin(); it != g[x].end(); it++) 11 #define db double 12 inline int rd(void) { 13 int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -1; 14 int x = 0; for (; isdigit(c); c = getchar()) x = x*10+c-‘0‘; return x*f; 15 } 16 17 const int N = 205; 18 const db INF = 1e20; 19 20 int n, m, S, T; 21 struct Edge { 22 int v; db x, y; 23 inline db d(db a) { return a * x + (1 - a) * y; } 24 }; 25 vector<Edge> g[N]; 26 27 inline void Init(int s, int t, db x, db y) { 28 g[s].push_back((Edge){t, x, y}); 29 g[t].push_back((Edge){s, x, y}); 30 } 31 32 int q[N], qh, qt; bool inq[N]; db dis[N]; 33 map<db, db> val; 34 35 db SPFA(db a) { 36 if (val.count(a)) return val[a]; 37 memset(q, 0, sizeof q), qh = qt = 0, memset(inq, false, sizeof inq), fill(dis + 1, dis + n + 1, INF); 38 q[++qt] = S, inq[S] = true, dis[S] = 0; 39 while (qh != qt) { 40 int x = q[qh = qh % n + 1]; 41 fore(it, x) 42 if (dis[x] + it->d(a) < dis[it->v]) { 43 dis[it->v] = dis[x] + it->d(a); 44 if (!inq[it->v]) 45 q[qt = qt % n + 1] = it->v, inq[it->v] = true; 46 } 47 inq[x] = false; 48 } 49 return val[a] = dis[T]; 50 } 51 52 inline db G(db L, db R) { return (R - L) * (SPFA(L) + SPFA(R) + 4 * SPFA((L+R)/2)) / 6; } 53 db Simpson(db L, db R) { 54 db M = (L+R)/2, RA = G(L, R), RL = G(L, M), RR = G(M, R); 55 return fabs(RA - RL - RR) < 1e-6 ? RL + RR : Simpson(L, M) + Simpson(M, R); 56 } 57 58 int main(void) { 59 #ifndef ONLINE_JUDGE 60 freopen("C.in", "r", stdin); 61 #endif 62 63 n = rd(), m = rd(), S = rd(), T = rd(); 64 F(i, 1, m) { 65 int s = rd(), t = rd(); 66 db x, y; scanf("%lf %lf", &x, &y); 67 Init(s, t, x, y); 68 } 69 printf("%0.10lf\n", Simpson(0, 1)); 70 71 return 0; 72 }
标签：path ace back cstring ctime pac struct print min
原文地址：http://www.cnblogs.com/Sdchr/p/7367312.html
