2011年 ACM/ICPC 北京赛区 A题 Qin shi huang's national road system

现场其实想到了找两点间最大边，但是一直在想DFS找的方法，期间我也想了想预处理的可行性，可是想岔道了，觉得不可行，结果今天用类似DP的预处理终于给过了，我擦了，铁牌第一名，要多点背有多点背。莫非是RP用光了，希望明年能好点吧。

Run IDProblem IDStatusTimeMemoryLanguageCodeSubmit AtUser

15470238Accepted216ms15956kbG++3153B2011-10-25 08:44:20maddoctor

这道题的解法其实有两种，另一种是先求一棵最小生成树，然后枚举每一条最小生成树上的边，拆掉后变成两个生成树，然后找两个集合中点权最大的两个连接起来，这是暴力贪心策略。而正解是北大教练讲的，先求一棵最小生成树，通过类似DP的预处理将每两点之间的最大边求出来，然后，一个二重循环枚举每对点，减掉最大边，添上点权就行。

/*ID: sdj22251PROG: subsetLANG: C++*/#include <iostream>#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <cctype>#include <string>#include <cstring>#include <cmath>#include <ctime>#define LOCA#define MAXN 10005#define INF 100000000#define eps 1e-11#define L(x) x<<1#define R(x) x<<1|1using namespace std;double d[1005][1005]; //存储两点间距离double dis[1005]; bool used[1005];int n;double mx[1005][1005];int near[1005]; //记录每次所加的边，如果near[i] = -1 就说明顶点i已经加到生成树中了。struct point{    double x, y;    int po;} p[1005];double distan(point x, point y){    return sqrt((x.x - y.x) * (x.x - y.x) + (x.y - y.y) * (x.y - y.y));}int main(){#ifdef LOCAL    freopen("d:/data.in","r",stdin);    freopen("d:/data.out","w",stdout);#endif    int T, i, j, n;    scanf("%d", &T);    while(T--)    {        scanf("%d", &n);        for(i = 1; i <= n; i++)            scanf("%lf%lf%d", &p[i].x, &p[i].y, &p[i].po);        for(i = 1; i <= n; i++)        {            for(j = 1; j <= n; j++)            {                d[i][j] = d[j][i] = distan(p[i], p[j]);                mx[i][j] = 0;            }        }        memset(used, false, sizeof(used));        for(i = 1; i <= n; i++)        {            dis[i] = d[1][i];            near[i] = 1;        }        used[1] = true;        double B = 0;        int A;        near[1] = -1;        for(i = 1; i < n; i++)        {            int mi = INF;            int v = -1;            for(j = 1; j <= n; j++)            {                if(near[j] != -1 && dis[j] < mi)                {                    v = j;                    mi = dis[j];                }            }            int pre;            if(v != -1)            {                pre = near[v];                B += d[v][near[v]];                //printf("%d %d ddd\n", near[v], v);                mx[near[v]][v] = mx[v][near[v]] = d[v][near[v]]; //邻接的点之间的最大边是自己                near[v] = -1;                for(j = 1; j <= n; j++)                {                    if(near[j] != -1 && d[v][j] < dis[j])                    {                        dis[j] = d[v][j];                        near[j] = v;                    }                }            }            for(j = 1; j <= n; j++)            {                if(near[j] == -1 && j != v)                {                    mx[j][v] = mx[v][j] = max(mx[j][pre], mx[pre][v]); //转移方程，代表j到v的最大边为j到pre和pre到v的最大边中的大的。                }            }        }        double ans = 0;        for(i = 1; i <= n; i++) //二重循环遍历        {            for(j = 1; j <= n; j++)            {                if(i == j) continue;                A = p[i].po + p[j].po;                if((double)A / (B - mx[i][j]) > ans)                    ans = (double)A / (B - mx[i][j]);            }        }        printf("%.2f\n", ans);    }    return 0;}