1. 题目描述
一个人沿着一条长度为n个链行走，给出了每秒钟由i到j的概率（$i,j \in [1,n]$）。求从1开始走到n个时间的期望。
2. 基本思路
显然是个DP。公式推导也相当容易。不妨设$dp[i], i \in [1,n]$表示由i到n的期望时间。
\begin{align}
dp[i] &= \Sigma_{j=1}^{n} p(i, j) (dp[j] + 1), &j<n\\
dp[i] &= 0 &i=n
\end{align}
显然这是一个n元方程组，可以高斯消元解。但是因为n很大，因此不能直接套用高斯消元。但是通过观察系数矩阵可以发现规律。
以$n=7, m=2$为例，$\times$表示$A_{ij}$不为0。
发现每个行向量最多包含$2m+1$个非零向量，即$[i-m, i+m]$。因此，在高斯消元的过程中，实际每次需要减掉的系数最多也就$2m+1$个。
因为$m \in [1,5]$，可以直接模拟非零系数的消元。因为$i-m$有可能小于0，因此，将$[i-m,i+m]$映射到$[0,2m]$的区间内，$A_{ii}$恰好映射到$P_{im}$。
最后可以生成几个小规模的n，与高斯消元对拍一下。
3. 代码
/* 4579 */
#include <iostream>
#include <sstream>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <deque>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <climits>
#include <cctype>
#include <cassert>
#include <functional>
#include <iterator>
#include <iomanip>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,1024000") #define sti set<int>
#define stpii set<pair<int, int> >
#define mpii map<int,int>
#define vi vector<int>
#define pii pair<int,int>
#define vpii vector<pair<int,int> >
#define rep(i, a, n) for (int i=a;i<n;++i)
#define per(i, a, n) for (int i=n-1;i>=a;--i)
#define clr clear
#define pb push_back
#define mp make_pair
#define fir first
#define sec second
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define lson l, mid, rt<<1
#define rson mid+1, r, rt<<1|1 const double eps = 1e-;
const int maxn = ;
const int maxm = ;
double g[maxn][maxm], p[maxn][maxm];
double v[maxn], x[maxn];
int C[maxn][];
int n, m; void solve() {
int i, j, k; memset(p, , sizeof(p));
for (i=; i<n; ++i) {
int tot = ;
double tmp = 0.0;
for (j=; j<=m; ++j)
tot += C[i][j]; for (j=; j<=m; ++j) {
if (i-j >= ) {
p[i][m-j] = 0.3 * C[i][j] / tot;
tmp += p[i][m-j];
}
if (i+j <= n) {
p[i][m+j] = 0.7 * C[i][j] / tot;
tmp += p[i][m+j];
}
}
p[i][m] = -tmp;
v[i] = -;
}
p[n][m] = ;
v[n] = ; memcpy(g[], p[], sizeof(p[]));
for (i=,k=; i<=n; ++i,++k) {
int l = max(k-m, );
int r = min(k+m, n);
for (j=i; j<=n&&j-k<=m; ++j) {
if (fabs(p[k][m]) < eps)
continue;
double t = p[j][k-j+m] / p[k][m];
for (int kk=k+; kk<=n&&kk-k<=m; ++kk)
p[j][kk-j+m] -= t * p[k][kk-k+m];
v[j] -= t * v[k];
} l = max(i-m, );
r = min(i+m, n);
for (j=l; j<=r; ++j)
g[i][j-i+m] = p[i][j-i+m];
} x[n] = ;
for (i=n-,k=n; i>; --i,--k){
for (j=i; j>&&k-j<=m; --j)
v[j] -= x[k] * g[j][k-j+m];
x[i] = v[i] / g[i][m];
} printf("%.2lf\n", x[]);
} int main() {
ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
freopen("data.in", "r", stdin);
freopen("data.out", "w", stdout);
#endif while (scanf("%d%d",&n,&m)!=EOF && (n||m)) {
rep(i, , n+)
rep(j, , m+)
scanf("%d", &C[i][j]);
solve();
} #ifndef ONLINE_JUDGE
printf("time = %d.\n", (int)clock());
#endif return ;
}
4. 数据生成器
import sys
import string
from random import randint def GenData(fileName):
with open(fileName, "w") as fout:
t = 10
for tt in xrange(t):
n = randint(1, 200)
m = randint(1, 5)
fout.write("%d %d\n" % (n, m))
L = [0] * m
for i in xrange(n):
for j in xrange(m):
L[j] = randint(1, 9)
fout.write(" ".join(map(str, L)) + "\n")
fout.write("0 0\n") def MovData(srcFileName, desFileName):
with open(srcFileName, "r") as fin:
lines = fin.readlines()
with open(desFileName, "w") as fout:
fout.write("".join(lines)) def CompData():
print "comp"
srcFileName = "F:\Qt_prj\hdoj\data.out"
desFileName = "F:\workspace\cpp_hdoj\data.out"
srcLines = []
desLines = []
with open(srcFileName, "r") as fin:
srcLines = fin.readlines()
with open(desFileName, "r") as fin:
desLines = fin.readlines()
n = min(len(srcLines), len(desLines))-1
for i in xrange(n):
ans2 = int(desLines[i])
ans1 = int(srcLines[i])
if ans1 > ans2:
print "%d: wrong" % i if __name__ == "__main__":
srcFileName = "F:\Qt_prj\hdoj\data.in"
desFileName = "F:\workspace\cpp_hdoj\data.in"
GenData(srcFileName)
MovData(srcFileName, desFileName)