POJ 2253 Frogger（最小生成树）

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题目链接：点击打开链接

题意：求从1到2的路径中， 使得最长路尽量小。

细节参见代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const ld eps = 1e-9, PI = 3.1415926535897932384626433832795;
const int mod = 1000000000 + 7;
const int INF = int(1e9);
// & 0x7FFFFFFF
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 200 + 10;
int T,n,m,cnt,p[maxn],kase=0;
double ans[maxn][maxn],x[maxn],y[maxn];
struct node {
    int a, b;
    double dist;
    node(int a=0, int b=0, double dist=0):a(a), b(b), dist(dist) {}
    bool operator < (const node& rhs) const {
        return dist < rhs.dist;
    }
}a[maxn*maxn];
vector<node> g[maxn];
int _find(int x) { return p[x] == x ? x : p[x] = _find(p[x]); }
void dfs(int u, int fa) {
    int len = g[u].size();
    for(int i=0;i<len;i++) {
        int v = g[u][i].a;
        if(v != fa) {
            ans[1][v] = max(ans[1][u], g[u][i].dist);
            dfs(v, u);
        }
    }
}
double solve() {
    for(int i=1;i<=n;i++) p[i] = i, g[i].clear();
    sort(a, a+cnt);
    for(int i=0;i<cnt;i++) {
        int x = _find(a[i].a);
        int y = _find(a[i].b);
        if(x != y) {
            p[x] = y;
            g[a[i].a].push_back(node(a[i].b, 0, a[i].dist));
            g[a[i].b].push_back(node(a[i].a, 0, a[i].dist));
        }
    }
    ans[1][1] = -1.0;
    dfs(1, 1);
    return ans[1][2];
}
int main() {
    while(~scanf("%d",&n) && n) {
        for(int i=1;i<=n;i++) scanf("%lf%lf",&x[i],&y[i]);
        cnt = 0;
        for(int i=1;i<=n;i++) {
            for(int j=i+1;j<=n;j++) {
                double cur = sqrt((x[i] - x[j])*(x[i] - x[j]) + (y[i] - y[j])*(y[i] - y[j]));
                a[cnt++] = node(i,j,cur);
            }
        }
        printf("Scenario #%d\n",++kase);
        printf("Frog Distance = %.3f\n\n",solve());
    }
    return 0;
}

POJ 2253 Frogger（最小生成树）

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原文地址：http://blog.csdn.net/weizhuwyzc000/article/details/50807480