C++博客-&豪-随笔分类-算法&ACM

用stl打spfa短了1k代码，慢了200ms

Sat, 03 Nov 2007 08:40:00 GMT

原来用stl的优先队列这么爽，比赛时候多用，heap太容易打错了，毕竟没ghost_wei那么bt(heap,就几行,都打烂了-_-)

pku3159:

#include <iostream>
#include <vector>
#include <queue>
using namespace std;

const int INF = 1 << 28;
const int MAXN = 30010;

struct PQNode {
    int u, key;
    //pq默认用<判断优先级,key大优先,若要key小优先,则加上!或<改成>即可
    friend bool operator<(const PQNode &a, const PQNode &b) { return !(a.key < b.key); }

};

int n, m;
vector<int> adjv[MAXN], adjw[MAXN];

int dijkstraPQ(int st, int en) {
    int i, v, w, dist[MAXN], chk[MAXN];
    priority_queue<PQNode> pq;
    PQNode tmp, cp;

    memset(chk, 0, sizeof(chk));
    for (i=0; i<n; i++) dist[i] = INF;

    dist[st] = 0;
    tmp.u = st; tmp.key = 0;
    pq.push(tmp);
    while (!pq.empty()) {
        cp = pq.top();
        pq.pop();
        if (cp.u == en) return dist[en];
        if (chk[cp.u]) continue;
        chk[cp.u] = 1;
        for (i=0; i<adjv[cp.u].size(); i++) {
            v = adjv[cp.u][i]; w = adjw[cp.u][i];
            if (!chk[v] && (dist[v]==INF || dist[v]>cp.key+w)) {
                dist[v] = cp.key+w;
                tmp.u = v; tmp.key = dist[v];
                pq.push(tmp);
            }
        }
    }
    return -1;
}

int main() {
    int i, j, k, u, v, w;
    freopen("input.txt", "r", stdin);
    scanf("%d%d", &n, &m);
    for (i=0; i<m; i++) {
        scanf("%d%d%d", &u, &v, &w);
        u--; v--;
        adjv[u].push_back(v);
        adjw[u].push_back(w);
    }
    printf("%d\n", dijkstraPQ(0, n-1));
    return 0;
}

豪 2007-11-03 16:40 发表评论

三次样条插值

Sat, 20 Oct 2007 05:07:00 GMT

具体推导看书<<数值分析>>
code：

#include <iostream>
using namespace std;

const int MAXN = 100;

int n;
double x[MAXN], y[MAXN]; //下标从0..n
double alph[MAXN], beta[MAXN], a[MAXN], b[MAXN];
double h[MAXN];
double m[MAXN]; //各点的一阶导数;

inline double sqr(double pa) {
    return pa * pa;
}

double sunc(double p, int i) {
    return (1 + 2 * (p - x[i]) / (x[i + 1] - x[i])) * sqr((p - x[i + 1]) / (x[i + 1] - x[i])) * y[i]
            + (1 + 2 * (p - x[i + 1]) / (x[i] - x[i + 1])) * sqr((p - x[i]) / (x[i + 1] - x[i])) * y[i + 1]
            + (p - x[i]) * sqr((p - x[i + 1]) / (x[i] - x[i + 1])) * m[i]
            + (p - x[i + 1]) * sqr((p - x[i]) / (x[i + 1] - x[i])) * m[i + 1];
}

int main() {
    int i, j;
    double xx;
    freopen("threeInsert.in", "r", stdin);
    scanf("%d", &n);
    for (i = 0; i <= n; i++) scanf("%lf%lf", &x[i], &y[i]);
    // scanf("%lf%lf", &m[0], &m[n]);
    for (i = 0; i <= n - 1; i++) h[i] = x[i + 1] - x[i];
    //第一种边界条件
    //alph[0] = 0; alph[n] = 1; beta[0] = 2 * m[0]; beta[n] = 2 * m[n];
    //第二种边界条件
    alph[0] = 1; alph[n] = 0; beta[0] = 3 * (y[1] - y[0]) / h[0]; beta[n] = 3 * (y[n] - y[n - 1] / h[n - 1]);
    for (i = 1; i <= n - 1; i++) {
        alph[i] = h[i - 1] / (h[i - 1] + h[i]);
        beta[i] = 3 * ((1 - alph[i]) * (y[i] - y[i - 1]) / h[i - 1] + alph[i] * (y[i + 1] - y[i]) / h[i]);
    }
    a[0] = - alph[0] / 2; b[0] = beta[0] / 2;
    for (i = 1; i <= n; i++) {
        a[i] = - alph[i] / (2 + (1 - alph[i]) * a[i - 1]);
        b[i] = (beta[i] - (1 - alph[i]) * b[i - 1]) / (2 + (1 - alph[i]) * a[i - 1]);
    }
    m[n + 1] = 0;
    for (i = n; i >= 0; i--) {
        m[i] = a[i] * m[i + 1] + b[i];
    }
    scanf("%lf", &xx);
    for (i = 0; i < n; i++) {
        if (xx >= x[i] && xx <= x[i + 1]) break;
    }
    printf("%lf\n", sunc(xx, i));
    return 0;
}

豪 2007-10-20 13:07 发表评论

龙贝格积分2457jlu

Fri, 19 Oct 2007 17:02:00 GMT

Sailboat

Problem H: Sailboat

In the sailboat race, the contestant is requested to along with the prearrange path. Sailing ship's power comes from wind power and contestant's manpower. The wind power can completely used.

In a competition, the contestants are requested to along with a 1/4 circles with radius R, the sailboat will goto east from south. During this process, the wind direction is straight from west to the east with constant speed and power.

In order to maintain the travel direction, the athlete must adjust the sail to the vertical angle from movement direction in any time.

If the speed of sailboat is proportional to the power at movement direction, the proportional factor is k. Supposes the wind power is f, the athlete manpower is h, please given the time of sailboat from the beginning to the end.

Input

The first line of each case consists of 4 double number, that is radius of path: R, wind power: f,athlete manpower: h and proportional factor:k. In order to avoid the floating point error, you needn't output the answer directly. The next line is a integer n, the following n lines gives a double value which is candidate answer.

Output

For each candidate of each case, Only "Yes" or "No" should be printed. Output "Yes" if the relative error to your answer is less than 3%, otherwise "No". For example, if the model answer is 100, and the candidate is 98 or 102, you should output "Yes". Output one blank line between neighboring case

Sample Input

1.0 2.0 1.0 1.0
2
0.35
0.76

Sample Output

No
Yes

Problem Source: provided by skywind

#include <iostream>
#include <cmath>
using namespace std;

const int MAXN = 100;
const double PI = acos(-1.0);

double R, F, H, K, ans;
int n, cas;

double func(double x) {
    return R / K / (H + F * cos(x));
}

double romberg(double a, double b, double EPS = 1e-6) {
    double t[MAXN][MAXN] = {0}, tmp;
    int i, j, k, k2, m, m4;
    t[0][0] = (func(a) + func(b)) * (b - a) / 2;
    k = 1; k2 = 1;
    while (1) {
        tmp  = 0;
        for (i = 1; i <= k2; i++) {
            tmp += func(a + (2 * i - 1) * (b - a) / (2 * k2));
        }
        t[0][k] = (t[0][k - 1] + tmp * (b - a) / k2) / 2;
        for (m = 1, m4=4; m <= k; m++, m4 *= 4) {
            t[m][k - m] = (m4 * t[m - 1][k - m + 1] - t[m - 1][k - m]) / (m4 - 1);
        }
        if (fabs(t[k][0] - t[k - 1][0]) < EPS) break;
        k++; k2 *= 2;
    }
    return t[k][0];
}

void solve() {
    double tmp;
    scanf("%lf", &tmp);
    if (fabs(tmp - ans) / ans < 0.03) printf("Yes\n");
    else printf("No\n");
}

int main() {
    freopen("2457.in", "r", stdin);
    while (scanf("%lf%lf%lf%lf%d", &R, &F, &H, &K, &n) != EOF) {
        if (cas) printf("\n");
        else cas++;
        ans = romberg(0, PI/2);
        while (n--) {
            solve();
        }
    }
    return 0;
}

豪 2007-10-20 01:02 发表评论

三分法求函数极值buaa1024

Thu, 18 Oct 2007 03:00:00 GMT

Easy Problem

Time limit:1000 ms Memory limit:65536 KB

Total Submit:1755 (462 users) Accepted Submit:366 (332 users)

Description

In this problem, you're to calculate the distance between a point P(x_p, y_p, z_p) and a segment (x₁, y₁, z₁) − (x₂, y₂, z₂), in a 3D space, i.e. the minimal distance from P to any point Q(x_q, y_q, z_q) on the segment (a segment is part of a line).

Input

The first line contains a single integer T (1 ≤ T ≤ 1000), the number of test cases. Each test case is a single line containing 9 integers x_p, y_p, z_p, x₁, y₁, z₁, x₂, y₂, z₂. These integers are all in [-1000,1000].

Output

For each test case, print the case number and the minimal distance, to two decimal places.

Sample Input

3
0 0 0 0 1 0 1 1 0
1 0 0 1 0 1 1 1 0
-1 -1 -1 0 1 0 -1 0 -1

Sample Output

Case 1: 1.00
Case 2: 0.71
Case 3: 1.00

Problem Source

The 32nd ACM-ICPC Beijing First Round Internet Contest

其实和二分差不多，划个函数曲线出来，分三段，比划一下就很容易理解了：）

#include <iostream>
#include <cmath>
using namespace std;

double dist(double l[], double r[]) {
    return sqrt((l[0]-r[0])*(l[0]-r[0])+(l[1]-r[1])*(l[1]-r[1])+(l[2]-r[2])*(l[2]-r[2]));
}

int main() {
   // freopen("1024.in", "r", stdin);
    int n, cas=0;
    double l[3], r[3], p[3], p1[3], p2[3], d1, d2;
    scanf("%d", &n);
    while (n--) {
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf%lf", &p[0], &p[1], &p[2], &l[0], &l[1], &l[2], &r[0], &r[1], &r[2]);
        while (dist(l, r) > 1e-4) {
            p1[0] = (l[0] + r[0]) / 2;
            p1[1] = (l[1] + r[1]) / 2;
            p1[2] = (l[2] + r[2]) / 2;
            p2[0] = (r[0] + p1[0]) / 2;
            p2[1] = (r[1] + p1[1]) / 2;
            p2[2] = (r[2] + p1[2]) / 2;
            d1 = dist(p1, p); d2 = dist(p2, p);
            if (d2 >= d1) {
                r[0] = p2[0]; r[1] = p2[1]; r[2] = p2[2];
            } else {
                l[0] = p1[0]; l[1] = p1[1]; l[2] = p1[2];
            }
        }
        printf("Case %d: %.2lf\n", ++cas, dist(p,l));
    }
}

豪 2007-10-18 11:00 发表评论

为了大作业报告好写点，重写了RBTree

Sun, 23 Sep 2007 13:34:00 GMT

扩展了一点，有兴趣的可以去看看，MFC去，sigh~
http://www.cppblog.com/qywyh/articles/32740.html

豪 2007-09-23 21:34 发表评论