hdu3976[Electric resistance] hdu5006[Resistance] hdu4870[Rating] 高斯消元

写了几个高斯消元的题，试一下模板

hdu3976

题意：给定50个节点，2000个电阻，求1到n的等效电阻

这题要用 节点电位法。

在1和n之间加一个1A的电流源，将n接地，这样求出的1号点电位数值上等于等效电阻。

#include <cstdio>#include <iostream>typedef long long ll;using namespace std;void in(int &x){    char ch; int minus = 0;    while (ch=getchar(), (ch<'0'||ch>'9') && ch!='-');    if (ch == '-') minus = 1, x = 0;    else x = ch-'0';    while (ch=getchar(), ch>='0'&&ch<='9') x = x*10+ch-'0';    if (minus) x = -x;}void in(ll &x){    char ch; int minus = 0;    while (ch=getchar(), (ch<'0'||ch>'9') && ch!='-');    if (ch == '-') minus = 1, x = 0;    else x = ch-'0';    while (ch=getchar(), ch>='0'&&ch<='9') x = x*10+ch-'0';    if (minus) x = -x;}void out(int x){    char hc[30];int len, minus=0;    if (x<0) minus = 1, x = -x;    len = 0; hc[len++] = x%10+'0';    while (x/=10) hc[len++] = x%10+'0';    if (minus) putchar('-');    for (int i=len-1; i>=0; i--) putchar(hc[i]);}void out(ll x){    char hc[30];int len, minus=0;    if (x<0) minus = 1, x = -x;    len = 0; hc[len++] = x%10+'0';    while (x/=10) hc[len++] = x%10+'0';    if (minus) putchar('-');    for (int i=len-1; i>=0; i--) putchar(hc[i]);}#include <cstring>#include <cmath>const int N = 50+5;const double eps = 1e-6;double g[N][N];int n, m;void solve(int n, double g[][N]){    int now = 0;    for (int c=0; c<n; c++){        int k = -1;        for (int i=now; i<n; i++) if (fabs(g[i][c])>eps)            if (k==-1 || fabs(g[i][c])>fabs(g[k][c])) k = i;        if (k == -1) continue;        if (k != now)        for (int j=c; j<=n; j++)            swap(g[now][j], g[k][j]);        for (int i=0; i<n; i++) if (i!=now && fabs(g[i][c])>eps){            double tmp = g[i][c]/g[now][c];            for (int j=c; j<=n; j++)                g[i][j] -= tmp*g[now][j];        }        now ++;    }    for (int i=0; i<n; i++) if (fabs(g[i][i])>eps)        g[i][n] /= g[i][i];}int main(){#ifndef ONLINE_JUDGE    freopen("input.txt", "r", stdin);#endif    int test;    in(test);    for (int T=1; T<=test; T++){        in(n); in(m);        memset(g, 0, sizeof(g));        int a, b, c;        for (int i=1; i<=m; i++){            in(a); in(b); in(c); if (!c) continue;            a--; b--; double tmp = 1.0/c;            g[a][a] += tmp;            g[b][b] += tmp;            g[a][b] -= tmp;            g[b][a] -= tmp;        }        for (int i=0; i<n; i++) g[i][n] = 0.0;        g[0][n] = 1.0; g[n-1][n] = -1.0;        solve(n, g);        printf("Case #%d: %.2lf\n", T, g[0][n]-g[n-1][n]);    }    return 0;}

hdu5006

题意不变，求s到t之间的等效电阻，但是最多有10000个点，电阻的值只能随机取0或1

两节点之间电阻如果为0说明可以看做一个点，所以实际上点的数量非常少。

另外如果s和t在电路中不联通，则方程无解。

#include <cstdio>#include <iostream>typedef long long ll;using namespace std;void in(int &x){    char ch; int minus = 0;    while (ch=getchar(), (ch<'0'||ch>'9') && ch!='-');    if (ch == '-') minus = 1, x = 0;    else x = ch-'0';    while (ch=getchar(), ch>='0'&&ch<='9') x = x*10+ch-'0';    if (minus) x = -x;}void in(ll &x){    char ch; int minus = 0;    while (ch=getchar(), (ch<'0'||ch>'9') && ch!='-');    if (ch == '-') minus = 1, x = 0;    else x = ch-'0';    while (ch=getchar(), ch>='0'&&ch<='9') x = x*10+ch-'0';    if (minus) x = -x;}void out(int x){    char hc[30];int len, minus=0;    if (x<0) minus = 1, x = -x;    len = 0; hc[len++] = x%10+'0';    while (x/=10) hc[len++] = x%10+'0';    if (minus) putchar('-');    for (int i=len-1; i>=0; i--) putchar(hc[i]);}void out(ll x){    char hc[30];int len, minus=0;    if (x<0) minus = 1, x = -x;    len = 0; hc[len++] = x%10+'0';    while (x/=10) hc[len++] = x%10+'0';    if (minus) putchar('-');    for (int i=len-1; i>=0; i--) putchar(hc[i]);}#include <cstring>#include <cmath>#include <cassert>const int N = 2500;const int M = 40000+10;const double eps = 1e-7;int n, m, st, ed;double g[N][N], ans[N];int fa[N*4+10], u[M], v[M], c[M], ms[M/4];int find(int x){    if (fa[x] == x) return x;    return fa[x] = find(fa[x]);}bool solve(int n, double g[][N], double ans[]){    int now = 0;    for (int c=0; c<n; c++){        int r = now;        for (int i=now; i<n; i++) if (fabs(g[i][c]) > fabs(g[r][c]))            r = i;        if (r != now)            for (int j=c; j<=n; j++) swap(g[r][j], g[now][j]);        if (fabs(g[now][c]) < eps) continue;        for (int i=0; i<n; i++) if (i != now && fabs(g[i][c])>eps){            double tmp = g[i][c]/g[now][c];            for (int j=c; j<=n; j++)                g[i][j] -= tmp*g[now][j];        }        now ++;    }    for (int j=0; j<n; j++)        for (int i=0; i<n; i++) if (fabs(g[i][j]) > eps){            ans[j] = g[i][n]/g[i][j];            break;        }    for (int i=0; i<n; i++) if (fabs(g[i][n])>eps){        int flag = 0;        for (int j=i; j<n; j++) if (fabs(g[i][j])>eps){            flag = 1; break;        }        if (!flag) return false;    }    return true;}int main(){#ifndef ONLINE_JUDGE    freopen("input.txt", "r", stdin);#endif    int test, start, finish;    in(test);    while (test--){        in(n); in(m); in(start); in(finish);        memset(ms, 0xff, sizeof(ms));        memset(g, 0, sizeof(g));        for (int i=1; i<=n; i++) fa[i] = i;        for (int i=1; i<=m; i++){            in(u[i]); in(v[i]); in(c[i]);            if (!c[i]){                int l = find(u[i]), r = find(v[i]);                if (l!=r) fa[l] = r;            }        }        int cnt = 0;        for (int i=1; i<=n; i++)            if (find(i) == i) ms[i] = cnt++;            else ms[i] = -1;        for (int i=1; i<=m; i++) if (c[i]){            int l = find(u[i]), r = find(v[i]);            if (l != r){                l = ms[l]; r = ms[r];                g[l][l] += 1; g[r][r] += 1;                g[l][r] -= 1; g[r][l] -= 1;            }        }        st = ms[find(start)]; ed = ms[find(finish)];        for (int i=0; i<cnt; i++) g[i][ed] = 0;        g[st][cnt] = 1; g[ed][cnt] = -1;        if (st == ed) printf("%.6lf\n", 0.0);//important        else        if (solve(cnt, g, ans)) printf("%.6lf\n", ans[st]);        else puts("inf");    }    return 0;}

hdu4870

概率dp，50分看成1，1000分就成了20

dp[i][j] 表示一个账号为i 另一个为j 时的期望数

转移方程 dp[i][j] = p*dp[i+1][j] + (1-p)*dp[i-2][j] + 1 

方程既跟后面的状态有关，又跟前面的状态有关，所以不能直接递推

这种题就要用高斯消元。

ps：写这题的时候有两个地方卡了很久

1、我用 q 代替 1-p ，导致了精度损失。

2、原来写高斯消元我会取绝对值最大的主元进行消元，据说可以减少精度损失。

但由于未知原因，这一题加了这个优化反而会损失精度。

#include <cstdio>#include <cstring>#include <cmath>#include <cassert>#include <algorithm>using namespace std;const double eps = 1e-9;const int LIM = 210;const int L = 20;int n;double p, q;double g[LIM+5][LIM+5];//, gg[LIM+5][LIM+5];int id[L+5][L+5], pre[LIM+5], next[LIM+5];double solve(int n, double g[][LIM+5], double ans[]){    int now = 0, r;    for (int c=0; c<n; c++){        int r = now;        for (int i=now; i<n; i++) if (fabs(g[i][c]) > eps && fabs(g[i][c]) < fabs(g[r][c]))            r = i;        for (r=now; r<n; r++) if (fabs(g[r][c])>eps) break;        if (r != now)            for (int j=c; j<=n; j++) swap(g[r][j], g[now][j]);        if (fabs(g[now][c]) < eps) continue;        for (int i=0; i<n; i++) if (i != now && fabs(g[i][c])>eps){            double tmp = g[i][c]/g[now][c];            for (int j=c; j<=n; j++)                g[i][j] -= tmp*g[now][j];        }        now ++;    }    return g[0][n]/g[0][0];}void calc(int r, int c){    int tr = max(r-2, 0), tc = c;    pre[id[r][c]] = id[tr][tc];    tr = r+1; if (tr>tc) swap(tr, tc);    if (tc == L) return;    next[id[r][c]] = id[tr][tc];}int main(){#ifndef ONLINE_JUDGE    freopen("input.txt", "r", stdin);#endif    memset(pre, 0xff, sizeof(pre));    memset(next, 0xff, sizeof(next));    for (int i=0, cnt = 0; i<L; i++)        for (int j=i; j<L; j++)            id[i][j] = cnt++;    for (int i=0; i<L; i++)        for (int j=i; j<L; j++){            calc(i, j);        }    while (scanf("%lf", &p) != EOF){        memset(g, 0, sizeof(g));        q = 1.0-p;        for (int i=0; i<L; i++)            for (int j=i; j<L; j++){                int k = id[i][j];                g[k][k] += 1;                if (pre[k] != -1) g[k][pre[k]] -= (1-p);                if (next[k] != -1) g[k][next[k]] -= p;                g[k][LIM] += 1;            }        double ans[LIM+10];        printf("%.6lf\n", solve(LIM, g, ans));    }    return 0;}

这一题还有另一种思路

原来状态方程是dp[i][j] = p*dp[i+1][j] + (1-p)*dp[i-2][j] + 1 ，跟前面、后面的状态都有关。

其实两个账号可以分开来看，一个账号你要达到20， 另一个则要达到19

所以可以用dp[i]表示从i到i+1所需期望步数，

转移方程为 dp[i] = p*1+(1-p)*(dp[i-2]+dp[i-1]+dp[i])，

这样就只涉及前面的状态了，可以直接o(n)的dp

#include <cstdio>using namespace std;const int N = 21;double p;double dp[N], s[N];int main(){    double p;    while (scanf("%lf", &p) != EOF){        dp[0] = 1/p;        dp[1] = 1/p/p;        for (int i=2; i<=20; i++){            dp[i] = (1/p-1)*(dp[i-2]+dp[i-1]) + 1/p;        }        s[0] = dp[0];        for (int i=1; i<=20; i++)            s[i] = s[i-1]+dp[i];        printf("%.6lf\n", s[19]+s[18]);    }    return 0;}