Hlg 树的直径.cpp

题意：
　　给出一棵树..求出一条路..使得别的点到这条路径的距离最长的最短..

输入：
　　　　n 表示树上有n个点..
　　　　u v w 表示u和v这两个相连的点的边权值为w

思路：
　　3次广搜..
　　第一次从任意一个点找最长的路径~这个路径的终点就是要求的路的终点..
　　第二次从终点开始找最长的路径..这时候找到的路径就是最后的路了..
　　然后从这个队列往外遍历相连的路..
　　这时候找出来的最长的距离就是要求的答案..

Tips：
　　因为加边的时候是双向边..
　　所以边的数组应该开两倍大..
　　pre 数组要注意~~~~

Code：

View Code

  1 #include <iostream>

  2 #include <cstring>

  3 #include <stdio.h>

  4 using namespace std;

  5 #define clr(x) memset(x, 0, sizeof(x))

  6 

  7 const int MAXN = 100010;

  8 

  9 struct Edge

 10 {

 11     int to;

 12     int next;

 13     int w;

 14 }edge[MAXN*2];

 15 int tot;

 16 int head[MAXN];

 17 

 18 void add(int s, int u, int w) {

 19     edge[tot].to = u;

 20     edge[tot].next = head[s];

 21     edge[tot].w = w;

 22     head[s] = tot++;

 23 

 24     edge[tot].to = s;

 25     edge[tot].next = head[u];

 26     edge[tot].w = w;

 27     head[u] = tot++;

 28 }

 29 

 30 struct Node

 31 {

 32     int num;

 33     int dis;

 34 }que[MAXN];

 35 

 36 bool vis[MAXN];

 37 int rd[MAXN];

 38 int bfs()

 39 {

 40     Node tmp, tmp1;

 41     int pre[MAXN];

 42     int tmps, ma = 0, s, e, ans = 0;

 43     clr(pre);

 44     int front = 0, rear = 0;

 45     clr(vis);

 46     tmp.num = 1;

 47     tmp.dis = 0;

 48     que[rear++] = tmp;

 49     vis[1] = true;

 50     while(front < rear) {

 51         tmp = que[front++];

 52         tmps = tmp.num;

 53         for(int i = head[tmps]; i; i = edge[i].next)

 54         if(!vis[edge[i].to]) {

 55             tmp1.num = edge[i].to;

 56             tmp1.dis = edge[i].w + tmp.dis;

 57             que[rear++] = tmp1;

 58             vis[edge[i].to] = true;

 59             if(ma < tmp1.dis) {

 60                 ma = tmp1.dis;

 61                 e = tmp1.num;

 62             }

 63         }

 64     }

 65 

 66     ma = front = rear = 0;

 67     clr(vis);

 68     tmp.num = e;

 69     tmp.dis = 0;

 70     que[rear++] = tmp;

 71     vis[e] = true;

 72     while(front < rear) {

 73         tmp = que[front++];

 74         tmps = tmp.num;

 75         for(int i = head[tmps]; i; i = edge[i].next)

 76         if(!vis[edge[i].to]) {

 77             tmp1.num = edge[i].to;

 78             tmp1.dis = edge[i].w + tmp.dis;

 79             vis[edge[i].to] = true;

 80             que[rear++] = tmp1;

 81             pre[tmp1.num] = tmps;

 82             if(ma < tmp1.dis) {

 83                 s = tmp1.num;

 84                 ma = tmp1.dis;

 85             }

 86         }

 87     }

 88 

 89     ma = front = rear = 0;

 90     clr(vis);

 91     tmp.num = s;

 92     tmp.dis = 0;

 93     que[rear++] = tmp;

 94 //printf("%d===>", s);

 95     while(pre[s]) {

 96 //printf("%d==>", pre[s]);

 97         tmp.num = pre[s];

 98         tmp.dis = 0;

 99         que[rear++] = tmp;

100         vis[s] = true;

101         s = pre[s];

102     }

103 //printf("%d", e);

104 //puts("");

105     vis[e] = true;

106     while(front < rear) {

107         tmp = que[front++];

108         tmps = tmp.num;

109         for(int i = head[tmps]; i; i = edge[i].next)

110         if(!vis[edge[i].to]) {

111             tmp1.num = edge[i].to;

112             tmp1.dis = edge[i].w + tmp.dis;

113             que[rear++] = tmp1;

114             vis[tmp1.num] = true;

115             if(tmp1.dis > ma) {

116                 ma = tmp1.dis;

117             }

118         }

119     }

120     return ma;

121 }

122 

123 

124 int main()

125 {

126     int i, j, k;

127     int n, a, b;

128     int u, v, w;

129     while(scanf("%d", &n) != EOF) {

130         if(n == 0) break;

131 

132         tot = 1;

133         clr(head);

134         for(i = 0; i < n-1; ++i) {

135             scanf("%d %d %d", &u, &v, &w);

136             add(u, v, w);

137         }

138 

139         printf("%d\n", bfs());

140     }

141     return 0;

142 }

 

题目链接：http://acm.hrbust.edu.cn/index.php?m=ProblemSet&a=showProblem&problem_id=1460