XTOJ1267 Highway 【最小生成树】

HighwayAccepted : 100      Submit : 336Time Limit : 4000 MS        Memory Limit : 65536 KBHighwayIn ICPCCamp there were n towns conveniently numbered with 1,2,…,n connected with (n−1) roads. The i-th road connecting towns ai and bi has length ci. It is guaranteed that any two cities reach each other using only roads.Bobo would like to build (n−1) highways so that any two towns reach each using only highways. Building a highway between towns x and y costs him δ(x,y) cents, where δ(x,y) is the length of the shortest path between towns x and y using roads.As Bobo is rich, he would like to find the most expensive way to build the (n−1) highways.InputThe input contains zero or more test cases and is terminated by end-of-file. For each test case:The first line contains an integer n. The i-th of the following (n−1) lines contains three integers ai, bi and ci.1≤n≤1051≤ai,bi≤n1≤ci≤108The number of test cases does not exceed 10.OutputFor each test case, output an integer which denotes the result.Sample Input51 2 21 3 12 4 23 5 151 2 21 4 13 4 14 5 2Sample Output1915SourceXTU OnlineJudge

这OJ接受%I64d %lld会报WA!!!

模仿kurskal 把边从大到小依次加进生成树里即可
随便一个点出发dfs ，得到最远的一个点为x
从x出发dfs，得到最远的点为y
那dis[x][y]是树中距离最远的2点距离
(x,y)也是数的直径
同时，树中任意一点k，距离其他点的最远距离必定<=max(dis[k][x],dis[k][y])
因此 求出x,y的dfs，模拟kurskal即可

#include<stdio.h>#include <iostream>#include<stdlib.h>#include<algorithm>#include<vector>#include<deque>#include<map>#include<set>#include<queue>#include<math.h>#include<string.h>#include<string>using namespace std;#define ll long long#define pii pair<int,int>const int inf = 1e9 + 7;const int N = 1e5+5;struct Edge{    int v,w,next;}edge[2*N];int head[N];inline void addEdge(int k,int u,int v,int w){    edge[k].v=v;    edge[k].w=w;    edge[k].next=head[u];    head[u]=k;}ll dis1[N];ll dis2[N];void dfs(int u,int fa,ll*dis){    for(int i=head[u];i!=-1;i=edge[i].next){        const int&v=edge[i].v;        const int&w=edge[i].w;        if(v!=fa){            dis[v]=dis[u]+w;            dfs(v,u,dis);        }    }}pii init(int n){    fill(dis1,dis1+n+1,0);    dfs(1,-1,dis1);    ll maxDis=0;    int idx=0;    for(int i=1;i<=n;++i){        if(dis1[i]>maxDis){            maxDis=dis1[i];            idx=i;        }    }    fill(dis1,dis1+n+1,0);    dfs(idx,-1,dis1);    int idx2=0;    maxDis=0;    for(int i=1;i<=n;++i){        if(dis1[i]>maxDis){            maxDis=dis1[i];            idx2=i;        }    }    fill(dis2,dis2+n+1,0);    dfs(idx2,-1,dis2);    return make_pair(idx,idx2);}ll slove(int n){    if(n==1){        return 0;    }    pii r=init(n);    ll ans=dis1[r.second];    for(int i=1;i<=n;++i){        if(i==r.first||i==r.second){            continue;        }        ans+=max(dis1[i],dis2[i]);    }    return ans;}int main(){    //freopen("/home/lu/Documents/r.txt","r",stdin);    //freopen("/home/lu/Documents/w.txt","w",stdout);    int n;    while(~scanf("%d",&n)){        fill(head,head+n+1,-1);        for(int i=0;i<n-1;++i){            int a,b,c;            scanf("%d%d%d",&a,&b,&c);            addEdge(2*i,a,b,c);            addEdge(2*i+1,b,a,c);        }        printf("%I64d\n",slove(n));    }    return 0;}