【HDOJ】4418 Time travel

1. 题目描述
K沿着$0,1,2,cdots,n-1,n-2,n-3,cdots,1,$的循环节不断地访问$[0, n-1]$个时光结点。某时刻，时光机故障，这导致K必须持续访问时间结点。故障发生在结点x处，方向为d，
在访问k个结点后时光机以概率$P_k%$的概率修复好，k不超过m。求当K最终访问结点Y时经过的时光结点的期望。

2. 基本思路
上述循环节包含包含$nn = 2n-2个$元素（因此，尤其需要特判n=1的情况，否则除0wa）。
通过x和方向d可以唯一的确定x在这个循环节中的位置。
设$E[i], i in [0, nn)$表示由结点i最终访问成功Y的期望。对期望进行推导
egin{align}
    E[0] &= (E[(0+1) \% nn]+1) imes P_1 + (E[(0+2) \% nn] + 2) imes P_2 + cdots (E[(0+m) \% nn] + m) imes P_m otag \
    E[1] &= (E[(1+1) \% nn]+1) imes P_1 + (E[(1+2) \% nn] + 2) imes P_2 + cdots (E[(1+m) \% nn] + m) imes P_m otag \
        &cdots otag \
    E[i] &= (E[(i+1) \% nn]+1) imes P_1 + (E[(i+2) \% nn] + 2) imes P_2 + cdots (E[(i+m) \% nn] + m) imes P_m \
    E[i] &= 0, quad if quad id[i]=y
end{align}
这里显然是一个nn元方程组，可以高斯消元解。
这题对精度有限制，因此最开始加入一个bfs，判定x能否走到y，这里一定要判定$P_j, j in [1,m]$是否近似于0。

3. 代码

/* 4418 */
#include <iostream>
#include <sstream>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <deque>
#include <bitset>
#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 int maxn = 220;
typedef double mat[maxn][maxn];

const double eps = 1e-8;
int n, nn, m, x, y, d;
int visit[maxn];
int id[maxn];
double P[maxn];
mat g;

bool gauss_elimination(int n) {
    int r;
    
    rep(i, 0, n) {
        r = i;
        rep(j, i+1, n) {
            if (fabs(g[j][i]) > fabs(g[r][i]))
                r = j;
        }
        
        if (r != i) {
            rep(j, 0, n+1)
                swap(g[r][j], g[i][j]);
        }
        
        if (fabs(g[i][i]) < eps)
            return false;
        
        rep(k, i+1, n) {
            double t = g[k][i] / g[i][i];
            rep(j, i+1, n+1)
                g[k][j] -= t * g[i][j];
        }
    }
    
    per(i, 0, n) {
        rep(j, i+1, n)
            g[i][n] -= g[i][j] * g[j][n];
        g[i][n] /= g[i][i];
    }
    
    return true;
}

int bfs(int bx) {
    queue<int> Q;
    int p = 0;
    bool ret = false;
    
    memset(visit, -1, sizeof(visit));
    visit[bx] = p++;
    Q.push(bx);
    
    while (!Q.empty()) {
        int u = Q.front();
        Q.pop();
        if (id[u] == y)
            ret = true;
        rep(j, 1, m+1) {
            if (fabs(P[j]) < eps)
                continue;
            int v = (u + j) % nn;
            if (visit[v] == -1) {
                visit[v] = p++;
                Q.push(v);
            }
        }
    }
    
    return ret ? p : 0;
}

void solve() {
    if (x == y) {
        puts("0.00");
        return ;
    }
    
    nn = n*2-2;
    
    rep(i, 0, n)
        id[i] = i;
    for (int i=n,j=n-2; i<nn; ++i,--j)
        id[i] = j;
    
    int bx;
    
    if (d == 0)
        bx = x;
    else if (d == 1)
        bx = nn - x;
    else
        bx = x;
    
    int vn = bfs(bx);
    if (!vn) {
        puts("Impossible !");
        return ;
    }
    
    memset(g, 0, sizeof(g));
    rep(i, 0, nn) {
        if (visit[i] == -1)
            continue;
        
        int p = visit[i];
        g[p][p] = 1;
        
        if (id[i] == y)
            continue;
        
        rep(j, 1, m+1) {
            int k = visit[(i + j) % nn];
            if (k == -1)
                continue;
            
            g[p][k] -= P[j];
            g[p][vn] += j * P[j];
        }
    }
    
    bool flag = gauss_elimination(vn);
    
    if (!flag || fabs(g[0][vn])<eps) {
        puts("Impossible !");
        return ;
    }
    
    printf("%.2lf
", g[0][vn]);
}

int main() {
    ios::sync_with_stdio(false);
    #ifndef ONLINE_JUDGE
        freopen("data.in", "r", stdin);
        freopen("data.out", "w", stdout);
    #endif
    
    int t;
    
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d%d%d%d", &n,&m,&y,&x,&d);
        rep(i, 1, m+1) {
            scanf("%lf", &P[i]);
            P[i] /= 100.0;
        }
        solve();
    }
    
    #ifndef ONLINE_JUDGE
        printf("time = %d.
", (int)clock());
    #endif
    
    return 0;
}

4. 数据生成器

import sys
import string
from random import randint

    
def GenData(fileName):
    with open(fileName, "w") as fout:
        t = 20
        for tt in xrange(t):
            n = randint(1, 100)
            m = randint(1, 100)
            y = randint(0, n-1)
            x = randint(0, n-1)
            if x==0 or x==n-1:
                d = -1
            else:
                d = randint(0, 1)
            fout.write("%d %d %d %d %d
" % (n, m, y, x, d))
            L = [0] * m
            tot = 0
            for i in xrange(m-1):
                x = randint(0, 100-tot)
                L[i] = x
                tot += x
            L[m-1] = 100 - tot
            fout.write(" ".join(map(str, L)) + "
")
                
        
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_prjhdojdata.out"
    desFileName = "F:workspacecpp_hdojdata.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_prjhdojdata.in"
    desFileName = "F:workspacecpp_hdojdata.in"
    GenData(srcFileName)
    MovData(srcFileName, desFileName)