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
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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

3 3 2
1 2 1 0
2 3 1 0
1 3 1 1
3 3 2
1 2 1 0
2 3 1 0
1 3 0 5

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 = 999999999;
struct Edge{
    int v,next;
    int c,d;
}edge[1000];
int head[100],tot;
int n,m,T;
void init(){
    memset(head,-1,sizeof(head));
    tot = 0;
}
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[100];
bool vis[100];
double F(double x){
    for(int i=1;i<=n;i++){
        dis[i] = INF;
        vis[i] = false;
    }
    dis[1] = 0;
    queue<int> q;
    q.push(1);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        vis[u] = false;
        for(int k = head[u];k!=-1;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)/2;
    return (F(a) + 4*F(c) + F(b))*(b-a)/6;
}
// 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)/2;
    double L = simpson(a,c) ,R = simpson(c,b);
    if(fabs(A-L-R)<=15*eps) return L + R +(A-L-R)/15;
    return asr(a,c,eps/2,L) + asr(c,b,eps/2,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=0;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(0,T,1e-6)/T;
        printf("%.8lf
",ans);
    }
    return 0;
}

csu 1742

1742: Integral Function

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 79  Solved: 27
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Description

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

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 ”,(int)(answer*10000)) ‘‘ ).

Sample Input

1 1
1 2
2 8

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)/2;
    return (F(a) + 4*F(c) + F(b))*(b-a)/6;
}
// 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)/2;
    double L = simpson(a,c) ,R = simpson(c,b);
    if(fabs(A-L-R)<=15*eps) return L + R +(A-L-R)/15;
    return asr(a,c,eps/2,L) + asr(c,b,eps/2,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
",(int)(asr(a,b,1e-6)*10000));
    }
    return 0;
}