UVALive 5864 Register Allocation 解题报告

题意：给出平面上n个点的坐标，给出起点和终点，找出一条路径使得起点到重点的经过的最长边最短，并输出这个最短的值。

解法：二分长度+bfs判断是否可以到达

//time 112ms#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <cstring>#include <set>#include <queue>const int maxn = 1005;using namespace std;struct Point{    int x,y;    void input(){scanf("%d%d",&x,&y);}}p[maxn];struct Edge{    int v,next;}edge[maxn*maxn];int n,s,t,en,head[maxn];int dist[maxn][maxn];int Dist(Point a,Point b){    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}void addedge(int u,int v){    edge[en].v = v;    edge[en].next = head[u];    head[u] = en++;}void build(int m){    en = 0;    memset(head,-1,sizeof(head));    for(int i = 1; i<=n; i++)        for(int j = i+1; j <= n; j++)            if(dist[i][j]<=m)                addedge(i,j),addedge(j,i);}bool bfs(int s,int t){    queue<int> q;    q.push(s);    bool vis[maxn]={0};    vis[s] = 1;    while(!q.empty())    {        int u = q.front();        q.pop();        for(int i = head[u]; i != -1; i=edge[i].next)        {            int v = edge[i].v;            if(v==t) return 1;            if(!vis[v])            {                vis[v] = 1;                q.push(v);            }        }    }    return 0;}int main(){    //freopen("/home/moor/Code/input","r",stdin);    while(scanf("%d",&n)&&n)    {        for(int i = 1; i<=n; i++)            p[i].input();        scanf("%d%d",&s,&t);        int l,r,m,ans;        l = r =0;        for(int i = 1; i<=n; i++)            for(int j = i+1; j <= n; j++)            {                dist[i][j] = dist[j][i] = Dist(p[i],p[j]);                r = max(r,dist[i][j]);            }        while(l<=r)        {            m = (l+r)>>1;            build(m);            if(bfs(s,t)) ans = m,r = m-1;            else l = m+1;        }        printf("%d\n",ans);    }    return 0;}