csu 1806 & csu 1742 (simpson公式+最短路)

1806: Toll

Time Limit: 5 Sec Memory Limit: 128 MB Special Judge
Submit: 256 Solved: 74
[Submit][Status][Web Board]

Description

In ICPCCamp, there are n cities and m unidirectional roads between cities. The i-th road goes from the a_i-th city to the b_i-th city. For each pair of cities u and v, there is at most one road from u to v.

As traffic in ICPCCamp is becoming heavier, toll of the roads also varies. At time t, one should pay (c_i⋅t+d_i) dollars to travel along the i-th road.

Bobo living in the 1-st city would like to go to the n-th city. He
wants to know the average money he must spend at least if he starts from
city 1 at t∈[0,T]. Note that since Bobo's car is super-fast, traveling on the roads costs him no time.

Formally, if f(t) is the minimum money he should pay from city 1 to city n at time t, Bobo would like to find

Input

The first line contains 3 integers n,m,T (2≤n≤10,1≤m≤n(n-1),1≤T≤10⁴).

The i-th of the following m lines contains 4 integers a_i,b_i,c_i,d_i (1≤a_i,b_i≤n,a_i≠b_i,0≤c_i,d_i≤10³).

It is guaranteed that Bobo is able to drive from city 1 to city n.

Output

A floating number denotes the answer. It will be considered correct if its absolute or relative error does not exceed 10^-6.

Sample Input

Sample Output

1.75000000

2.00000000

这东西实在太好用了,可惜省赛不会...simpson公式就是求定积分用的,这题的F函数就是在时间点为 t 时从 1点到n点的最小花费.

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

const double INF = ;

struct Edge{

    int v,next;

    int c,d;

}edge[];

int head[],tot;

int n,m,T;

void init(){

    memset(head,-,sizeof(head));

    tot = ;

}

void addEdge(int u,int v,int c,int d,int &k){

    edge[k].v = v,edge[k].c = c,edge[k].d = d,edge[k].next = head[u],head[u] = k++;

}

double dis[];

bool vis[];

double F(double x){

    for(int i=;i<=n;i++){

        dis[i] = INF;

        vis[i] = false;

    }

    dis[] = ;

    queue<int> q;

    q.push();

    while(!q.empty()){

        int u = q.front();

        q.pop();

        vis[u] = false;

        for(int k = head[u];k!=-;k = edge[k].next){

            int v = edge[k].v,c = edge[k].c,d = edge[k].d;

            double t = x*c+d;

            if(dis[v]>dis[u]+t){

                dis[v] = dis[u]+t;

                if(!vis[v]){

                    vis[v] = true;

                    q.push(v);

                }

            }

        }

    }

    double ret = dis[n];

    return ret;

}

// 三点simpson法。这里要求F是一个全局函数

double simpson(double a,double b){

    double c =  a+(b-a)/;

    return (F(a) + *F(c) + F(b))*(b-a)/;

}

// 自适应Simpson公式（递归过程）。已知整个区间[a,b]上的三点simpson值A

double asr(double a , double b ,double eps ,double A){

    double c = a+ (b-a)/;

    double L = simpson(a,c) ,R = simpson(c,b);

    if(fabs(A-L-R)<=*eps) return L + R +(A-L-R)/;

    return asr(a,c,eps/,L) + asr(c,b,eps/,R);

}

// 自适应Simpson公式（主过程）

double asr(double a, double b, double eps) {

  return asr(a, b, eps, simpson(a, b));

}

int main()

{

    while(scanf("%d%d%d",&n,&m,&T)!=EOF){

        init();

        for(int i=;i<m;i++){

            int u,v,c,d;

            scanf("%d%d%d%d",&u,&v,&c,&d);

            addEdge(u,v,c,d,tot);

        }

        double ans = asr(,T,1e-)/T;

        printf("%.8lf\n",ans);

    }

    return ;

}

csu 1742

1742: Integral Function

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 79 Solved: 27
[Submit][Status][Web Board]

Description

In mathematics, several function are unable to integral. For example:

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But you can get the answer by computer.

Input

There are no more than T (T<=30) cases. Each case include two integer a, b (0<a <= b<=10).

Output

Each case output an answer.

(Please output the answer by ‘‘ printf (“%d\n”,(int)(answer*10000)) ‘‘ ).

Sample Input

Sample Output

0

6593

-312

这题更好用..直接带进去算

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

double F(double x){

    return sin(x)/x;

}

// 三点simpson法。这里要求F是一个全局函数

double simpson(double a,double b){

    double c =  a+(b-a)/;

    return (F(a) + *F(c) + F(b))*(b-a)/;

}

// 自适应Simpson公式（递归过程）。已知整个区间[a,b]上的三点simpson值A

double asr(double a , double b ,double eps ,double A){

    double c = a+ (b-a)/;

    double L = simpson(a,c) ,R = simpson(c,b);

    if(fabs(A-L-R)<=*eps) return L + R +(A-L-R)/;

    return asr(a,c,eps/,L) + asr(c,b,eps/,R);

}

// 自适应Simpson公式（主过程）

double asr(double a, double b, double eps) {

  return asr(a, b, eps, simpson(a, b));

}

int main()

{

    double a,b;

    while(scanf("%lf%lf",&a,&b)!=EOF){

        printf("%d\n",(int)(asr(a,b,1e-)*));

    }

    return ;

}