ACM-ICPC Asia Training League 暑假第一阶段第五场 BCDI

B  求两个点到圆的切点的距离加上两个切点的弧长

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5+10;
double x,y,gx,gy,xa,xb,xc,ya,yb,yc;
double ans1[4], ans2[4];
double dis(double x1, double y1, double x2, double y2) {
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

int main() {
    double x, y, gx, gy, x1, y1, r1, x2, y2, r2;
    cin >> x >> y >> gx >> gy >> x1 >> y1 >> r1 >> x2 >> y2 >> r2;
    double xx1 = dis(x,y,x2,y2);
    double xx2 = dis(gx,gy,x2,y2);
    double ans1 = sqrt(xx1*xx1-r2*r2);
    double ans2 = sqrt(xx2*xx2-r2*r2);
    double xx3 = dis(x,y,gx,gy);
    double a1 = acos(r2/xx1);
    double a2 = acos(r2/xx2);
    double a3 = acos((xx1*xx1+xx2*xx2-xx3*xx3)/(2*xx1*xx2));
    double ans3 = (a3-a2-a1)*r2;
    printf("%.10f
",ans1+ans2+ans3);
    return 0;
}

C 滑雪问题，m条路径，从高处往地处滑，求能滑的最长路径是多少，从低往高处求，dfs可以过。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e3+10;
typedef pair<int,int> P;
vector<P> vs[N];
int d[N];
void dfs(int v) {
    for(auto p : vs[v]) {
        int u = p.first, w = p.second;
        if(d[u] < d[v] + w) {
            d[u] = d[v] + w;
            dfs(u);
        }
    }
}
int main() {
    int n, m;
    cin >> n >> m;
    for(int i = 0; i < m; i ++) {
        int u, v, w;
        cin >> u >> v >> w;
        vs[v].push_back(P(u,w));
    }
    for(int i = 1; i <= n; i ++) {
        dfs(i);
    }
    int MAX = 0;
    for(int i = 1; i <= n; i ++) MAX = max(MAX, d[i]);
    printf("%d
",MAX);
    return 0;
}

D n个数，求有多少中由n+1个数组成的

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
int n, x, MIN1 = INF, MIN2 = -INF;
int main() {
    cin >> n;
    int tmp = 0;
    for(int i = 1; i <= n; i ++) {
        cin >> x;
        tmp = x - tmp;
        if(i&1) MIN1 = min(MIN1, tmp);
        else MIN2 = max(MIN2,-tmp);
    }
    if(MIN2>MIN1) printf("0
");
    else if(MIN2 < 0 && MIN1 < 0) printf("0
");
    else if(MIN2 < 0 && MIN1 >= 0) printf("%d
",MIN1+1);
    else printf("%d
",MIN1-MIN2+1);
    return 0;
}

I 前面的等级大于后面的，只要前面的赢了后面的就一点赢，全部平均也算英雄赢。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5+10;
int n;
int a[N], b[N];
int main() {
    int x;
    cin >> n;
    for(int i = 0; i < n; i ++) cin >> a[i];
    for(int i = 0; i < n; i ++) cin >> b[i];
    int id = b[0] - a[0];
    if(id < 0) return 0*printf("0
");
    for(int i = 1; i < n; i ++) {
        if(b[i] - a[i] > id) {
            return 0*printf("%d
",id+1);
        } else if(b[i] - a[i] < id) {
            return 0*printf("%d
",id);
        }
    }
    printf("%d
",id);
    return 0;
}