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

2014-11-15 16:21 267 查看
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

;
int n, m;
void solve(int n, double g[]
){
    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]
 /= 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] += tmp;
            g[a][b] -= tmp;
            g[b][a] -= tmp;
        }
        for (int i=0; i<n; i++) g[i]
 = 0.0;
        g[0]
 = 1.0; g[n-1]
 = -1.0;
        solve(n, g);
        printf("Case #%d: %.2lf\n", T, g[0]
-g[n-1]
);
    }
    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

, ans
;
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[]
, 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]
/g[i][j];
            break;
        }
    for (int i=0; i<n; i++) if (fabs(g[i]
)>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

[b]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]
/g[0][0];
}
void calc(int r, int c){
    int tr = max(r-2, 0), tc = c;
    pre[id[r][c]] = id[tc];
    tr = r+1; if (tr>tc) swap(tr, tc);
    if (tc == L) return;
    next[id[r][c]] = id[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
, s
;
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;
}
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