The 2013 ACM-ICPC Asia Changsha Regional Contest(2013区域赛练习)

链接：http://acm.zju.edu.cn/onlinejudge/showProblems.do?contestId=1&pageNumber=28

比赛过了A， C ，G， H， J，K

A，J，K被小伙伴秒了，K题比较烦的搜索

H题二分就可以了，注意层数是F

C题几何题，是我A的，WA了2发，原来少了一种情况，还让队友帮忙检查，真心弱，高中解直角三角形题

G题先用havel算法构造出一个解，然后枚举4个不同点i,j,k,u，满足<i,j> <k,u>边存在 <i,k> <j,u>边不存在，然后把边<i,j> <k,u> 换成边 <i,k> <j,u> 就出现另外一组解了， 比赛的时候我们用随机数多次havel也过了。

D，I题貌似可以做的样子。

贴一下C，G的代码

C

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <iostream>using namespace std;const double eps = 1e-8;struct Point {    double x, y;    Point(double x, double y):    x(x), y(y){}    Point(){}    double len() {        return sqrt(x*x+y*y);    }    Point operator+(const Point t) {        return Point(x+t.x, y+t.y);    }    Point operator-(const Point t) {        return Point (x-t.x, y-t.y);    }    Point operator*(const double t) {        return Point(x*t, y*t);    }    Point operator/(const double t) {        return Point(x/t, y/t);    }    void in() {        scanf("%lf%lf", &x, &y);    }}o, v, tp;double cross(Point a, Point b) {    return a.x*b.y-a.y*b.x;}double dot(Point a, Point b) {    return a.x*b.x+a.y*b.y;}double r1, r2, r3;int main() {    int i, j;    while( ~scanf("%lf%lf%lf", &r1, &r2, &r3)) {        o.in(); v.in();        double d = fabs(cross(v, o))/v.len();        if(dot(v*(-1), o) < eps ||d-r2-r3 > eps) {            printf("%.6f\n", 0.0);            continue;        }        if(d-r1-r3 > eps) {            double d1 = r2+r3;            double sum = sqrt(d1*d1-d*d);            printf("%.6f\n", sum*2/v.len());            continue;        }        double d1 = r2+r3, d2 = r1+r3;        double sum  = sqrt(d1*d1-d*d)-sqrt(d2*d2-d*d);        printf("%.6f\n", sum*2/v.len());    }    return 0;}/*5 20 1 0 100 0 -15 20 1 30 15 -1 01 2 2 10 0 0 10*/

G

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <iostream>#include <vector>#include <cstdlib>#include <ctime>using namespace std;struct node {int d, v;bool operator<(const node &t) const {return d > t.d || (d==t.d && v > t.v);}node(int d, int v) :d(d), v(v) {}node() {}} p[104];int d[104];bool g[104][104];int n;bool havel() {int i, j;for(i = 1; i <= n; i++)p[i] = node(d[i], i);for(i = 1; i <= n; i++) {sort(p + i, p + n + 1);j = i + 1;while(p[i].d > 0 && j <= n) {if(p[j].d > 0&&p[i].d>0) {                int u=p[i].v;                int v=p[j].v;g[u][v] = g[v][u] = true;p[i].d--;p[j].d--;                j++;}else return 0;}if(p[i].d > 0) return 0;}return 1;}int sum;void out(bool g[][104]) {vector<pair<int, int> > v;int i, j;for(i = 1 ;i <= n; i++)for(j = i+1; j <= n; j++) if(g[i][j])v.push_back(make_pair(i, j));printf("%d %d\n", n, sum);for(i = 0; i < v.size(); i++) {if(i) printf(" ");printf("%d", v[i].first);}puts("");for(i = 0; i < v.size(); i++) {if(i) printf(" ");printf("%d", v[i].second);}puts("");}int main() {int i, j, k, u;while(~scanf("%d", &n)) {sum = 0;for(i = 1; i <= n; i++) {scanf("%d", &d[i]);sum += d[i];}sum>>=1;memset(g, false, sizeof(g));bool flag = havel();if(!flag) {puts("IMPOSSIBLE");continue;}int ii = -1, jj, kk, uu;for(i = 1; i <= n; i++)for(j = 1; j <= n; j++) if(g[i][j] && i != j)for(k = 1; k <= n; k++) if(!g[j][k] && k != i && k != j)for(u = 1; u <= n; u++) if(!g[i][u] && g[k][u] && u != i && u != j && u != k) {ii = i, jj = j, kk = k, uu = u;}if(ii == -1) {puts("UNIQUE");out(g);continue;}else {puts("MULTIPLE");out(g);//cerr << ii <<" " << jj << " " << kk << " "<< uu << "~~~~" << endl;g[ii][jj] = g[jj][ii] = false;g[kk][uu] = g[uu][kk] = false;g[ii][uu] = g[uu][ii] = true;g[jj][kk] = g[kk][jj] = true;out(g);}}return 0;}