THU-2016-3-1-TSP

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ACM模版

描述

题解

树归问题，名字虽然叫TSP，却和TSP没有啥关系，一道典型的树归问题。当然，方法不止一种，还可以先用拓扑排序预处理一下。

代码

#include <cstdio>#define MAX_N 100005#define MAX_M 1000005#define max(a, b) ((a > b) ? (a) : (b))using namespace std;const int INF = 0x3f3f3f3f;struct edge{    int to, w, next;    edge()    {        w = to = next = 0;    }};edge D[MAX_M];  //  邻接表int head[MAX_N];//  邻接表头int dp[MAX_M];  //  dp[i]:第i条边为开始的最长路int num[MAX_N]; //  num[i]:以i为起点的最长路int cnt = 1;//  往邻接表中添加边void add(int u, int v, int w){    D[cnt].to = v;    D[cnt].w = w;    D[cnt].next = head[u];    head[u] = cnt++;}//  树归部分int DFS(int s){    int maxn = 0;    for (int i = head[s]; i; i = D[i].next)    {        if (dp[i] == -1)        {            dp[i] = DFS(D[i].to) + D[i].w;        }        maxn = max(maxn, dp[i]);    }    return maxn;}//  从起点开始查找最长路void find(int s, int w){    printf("%d\t", s);    int minx = INF, W = 0;    for (int i = head[s]; i; i = D[i].next)    {        //  保证在最长路上并且字典序最小        if (dp[i] == w && minx > D[i].to)        {            W = w - D[i].w;            minx = D[i].to;        }    }    if (minx != INF)    {        find(minx, W);    }    return ;}int main(){    int n, m;    scanf("%d%d", &n, &m);    for (int i = 0; i <= m; i++)    {        dp[i] = -1;    }    int a, b, p;    for (int i = 0; i < m; i++)    {        scanf("%d%d%d", &a, &b, &p);        add(a, b, p);    }    int maxn = -1, temp = 0;    for (int i = 1; i <= n; i++)    {        num[i] = DFS(i);        if (num[i] > maxn)        {            maxn = num[i];            temp = i;        }    }    find(temp, maxn);    printf("\n");    return 0;}

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