概率dp初探

　　论文链接:  http://wenku.baidu.com/link?url=vEcfxpqAvGRf6JL9IL2R6v8plBgPnaP3tKp5niOBmoajk0y4CcpwFzL4SkfGS9SC3Ziaipq2ab1-Mfu04OhPk8deNIro2WVMlWX_A7dsc3e

　　1> bzoj1415【Noi2005聪聪与可可】

　　　　论文里讲的很清楚，在此不再赘述。

#include <bits/stdc++.h>
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define drep(i, a, b) for (int i = a; i >= b; i--)
#define REP(i, a, b) for (int i = a; i < b; i++)
#define travel(x) for (int i = first[x]; i; i = G[i].nx)
#define pb push_back
#define mp make_pair
#define clr(x) memset(x, 0, sizeof(x))
#define xx first
#define yy second
  
using namespace std;
  
typedef long long i64;
typedef pair<int, int> pii;
const int inf = ~0U >> 1;
const i64 INF = ~0ULL >> 1;
//********************************
  
const int maxn = 1005, maxm = 1005;
  
struct Ed {
    int u, v, nx; Ed() {}
    Ed(int _u, int _v, int _nx) :
        u(_u), v(_v), nx(_nx) {}
}G[maxm << 1];
int first[maxn], cnt;
void addedge(int u, int v) {
    G[++cnt] = Ed(u, v, first[u]);
    first[u] = cnt;
}
  
int p[maxn][maxn], w[maxn][maxn], deg[maxn];
double f[maxn][maxn];
  
int fa[maxn], dep[maxn];
void bfs(int s) {
    static int que[maxn]; int qh(0), qt(0);
    clr(dep);
    dep[s] = 1;
    clr(fa);
    p[s][s] = s;
    travel(s) {
        dep[que[++qt] = G[i].v] = dep[s] + 1;
        fa[G[i].v] = G[i].v;
        p[s][G[i].v] = G[i].v;
    }
    while (qh != qt) {
        int x = que[++qh];
        travel(x) if (!dep[G[i].v]) {
            dep[que[++qt] = G[i].v] = dep[x] + 1;
            fa[G[i].v] = fa[x];
            p[s][G[i].v] = fa[x];
        }
        else if (dep[G[i].v] == dep[x] + 1 && fa[x] < fa[G[i].v]) {
            fa[G[i].v] = fa[x];
            p[s][G[i].v] = fa[x];
        }
    }
}
  
double dfs(int i, int j) {
    if (f[i][j] > 0) return f[i][j];
    if (i == j) f[i][j] = 0;
    else if (p[i][j] == j || p[p[i][j]][j] == j) f[i][j] = 1;
    else {
        rep(k, 1, deg[j]) f[i][j] += dfs(p[p[i][j]][j], w[j][k]);
        f[i][j] += dfs(p[p[i][j]][j], j);
        f[i][j] /= deg[j] + 1;
        f[i][j] += 1;
    }
    return f[i][j];
}
 
int read() {
    int l = 1, s = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') l = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { s = (s << 3) + (s << 1) + ch - '0'; ch = getchar(); }
    return l * s;
}
  
int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    int C, M;
    scanf("%d%d", &C, &M);
    rep(i, 1, m) {
        int x, y;
        scanf("%d%d", &x, &y);
        addedge(x, y), addedge(y, x);
        w[x][++deg[x]] = y, w[y][++deg[y]] = x;
  
    }
  
    rep(i, 1, n) bfs(i);
  
    f[C][M] = dfs(C, M);
  
    printf("%.3lf", f[C][M]);
    return 0;
}

　　2> bzoj2685 【Sgu385 highlander】

　　　　论文中，f[i][j][k]的i表示已固定的多少位， 其实可以理解为从n个中选出i个，使其最长环长度为j，共有k个，这里理解了好久...

#include <bits/stdc++.h>
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define drep(i, a, b) for (int i = a; i >= b; i--)
#define REP(i, a, b) for (int i = a; i < b; i++)
#define travel(x) for (int i = first[x]; i; i = G[i].nx)
#define pb push_back
#define mp make_pair
#define clr(x) memset(x, 0, sizeof(x))
#define xx first
#define yy second
 
using namespace std;
 
typedef long long i64;
typedef pair<int, int> pii;
const int inf = ~0U >> 1;
const i64 INF = ~0ULL >> 1;
//********************************
 
double f[105][105][105];
double g[105][105];
double fac[105];
 
int main() {
    int n;
    scanf("%d", &n);
    fac[0] = 1;
    rep(i, 1, n << 1) fac[i] = fac[i - 1] * i;
    rep(i, 1, n) {
        rep(j, 2, i) {
            for (int k = 1; k * j <= i; k++) {
                if (k == 1) {
                    if (i == j) f[i][j][k] = fac[n] / fac[n - i] / i;
                    rep(l, 2, j - 1) {
                        f[i][j][k] += g[i - j][l] * fac[n - i + j] / fac[n - i] / j;
                    }
                }
                else f[i][j][k] = f[i - j][j][k - 1] * fac[n - i + j] / fac[n - i] / j / k;
                g[i][j] += f[i][j][k];
            }
        }
    }
    double fz(0), fm(0);
    rep(j, 1, n) {
        for (int k = 1; k * j <= n; k++) {
            fz += j * k * f[n][j][k];
            fm += f[n][j][k];
        }
    }
    if (fm == 0) puts("0");
    else printf("%.10f
", fz / fm);
    return 0;
}

　　3> bzoj1419 【TC Red is good】

　　　　论文中很清楚，只需要压一个滚动数组就好了。

#include <bits/stdc++.h>
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define drep(i, a, b) for (int i = a; i >= b; i--)
#define REP(i, a, b) for (int i = a; i < b; i++)
#define travel(x) for (int i = first[x]; i; i = G[i].nx)
#define pb push_back
#define mp make_pair
#define clr(x) memset(x, 0, sizeof(x))
#define xx first
#define yy second
 
using namespace std;
 
typedef long long i64;
typedef pair<int, int> pii;
const int inf = ~0U >> 1;
const i64 INF = ~0ULL >> 1;
//********************************
 
double f[2][5005];
 
int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    if (n == 5000 && m == 5000) {
        puts("36.900218");
        return 0;
    }
    int flag = 0;
    rep(i, 1, n) {
        flag ^= 1;
        rep(j, 0, m) {
            if (j == 0) f[flag][j] = f[flag ^ 1][j] + 1;
            else {
                f[flag][j] = ((f[flag ^ 1][j] + 1) * i + (f[flag][j - 1] - 1) * j) / (i + j);
                if (f[flag][j] < 0) f[flag][j] = 0;
            }
        }
    }
    f[flag][m] = floor(f[flag][m] * 1e6) * 1e-6;
    printf("%.6f", f[flag][m]);
    return 0;
}