[POJ 3268 Silver Cow Party ]Dijkstra

知识点：Dijkstra

1. 题目链接

[POJ 3268 Silver Cow Party ]

2. 题意描述

有n个顶点，m条边的有向图。求图中所有，从该节点出发到点x，然后从x回到该节点的最短路最长有多长。
(1≤N≤1000,1≤M≤100,000)

3. 解题思路

原图从x开始用dijkstra求一次最短路。然后将原图反向，再从x开始dijkstra求一次最短路。两次最短路累加求和，取最大值就是答案。

4. 实现代码

// C#ifndef _GLIBCXX_NO_ASSERT#include <cassert>#endif#include <cctype>#include <cerrno>#include <cfloat>#include <ciso646>#include <climits>#include <clocale>#include <cmath>#include <csetjmp>#include <csignal>#include <cstdarg>#include <cstddef>#include <cstdio>#include <cstdlib>#include <cstring>#include <ctime>#if __cplusplus >= 201103L#include <ccomplex>#include <cfenv>#include <cinttypes>#include <cstdalign>#include <cstdbool>#include <cstdint>#include <ctgmath>#include <cwchar>#include <cwctype>#endif// C++#include <algorithm>#include <bitset>#include <complex>#include <deque>#include <exception>#include <fstream>#include <functional>#include <iomanip>#include <ios>#include <iosfwd>#include <iostream>#include <istream>#include <iterator>#include <limits>#include <list>#include <locale>#include <map>#include <memory>#include <new>#include <numeric>#include <ostream>#include <queue>#include <set>#include <sstream>#include <stack>#include <stdexcept>#include <streambuf>#include <string>#include <typeinfo>#include <utility>#include <valarray>#include <vector>#if __cplusplus >= 201103L#include <array>#include <atomic>#include <chrono>#include <condition_variable>#include <forward_list>#include <future>#include <initializer_list>#include <mutex>#include <random>#include <ratio>#include <regex>#include <scoped_allocator>#include <system_error>#include <thread>#include <tuple>#include <typeindex>#include <type_traits>#include <unordered_map>#include <unordered_set>#endifusing namespace std;typedef long long LL;typedef long double LB;typedef pair<int, int> PII;typedef pair<LL, LL> PLL;const int INF = 0x3f3f3f3f;const LL INFL = 0x3f3f3f3f3f3f3f3fLL;const LB eps = 1e-8;const int MAXN = 1000 + 5;const int MAXM = 100000 + 5;template <typename T>inline bool scan_d (T &ret) {    char c;    int sgn;    if (c = getchar(), c == EOF) return 0; //EOF    while (c != '-' && (c < '0' || c > '9') ) c = getchar();    sgn = (c == '-') ? -1 : 1;    ret = (c == '-') ? 0 : (c - '0');    while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');    ret *= sgn;    return 1;}template<typename T>void print(T x) {    static char s[33], *s1; s1 = s;    if (!x) *s1++ = '0';    if (x < 0) putchar('-'), x = -x;    while(x) *s1++ = (x % 10 + '0'), x /= 10;    while(s1-- != s) putchar(*s1);}template<typename T> void println(T x) { print(x); putchar('\n');}int n, m, x;struct Dijkstra {    struct Edge {        int v, w, next;        Edge() {}        Edge (int v, int w, int next) : v (v), w (w), next (next) {}    } edges[MAXM];    struct QNode {        int u, w;        QNode() {}        QNode (int u, int w) : u (u), w (w) {}        bool operator > (const QNode& e) const {            return w > e.w;        }    } cur;    int head[MAXN], tot;    int dist[MAXN];    bool vis[MAXN];    priority_queue<QNode, vector<QNode>, greater<QNode> > Q;    void init() {        tot = 0;        memset(head, -1, sizeof (head) );        memset(dist, 0x3f, sizeof (dist) );        memset(vis, false, sizeof (vis) );    }    void add_edge(int u, int v, int w) {        edges[tot] = Edge (v, w, head[u]);        head[u] = tot ++;    }    void run(int src) {        int u, v, w;        Q.push(QNode (src, dist[src] = 0) );        while(!Q.empty() ) {            cur = Q.top();            Q.pop();            u = cur.u;            if(vis[u]) continue;            vis[u] = true;            for(int i = head[u]; ~i; i = edges[i].next) {                v = edges[i].v, w = edges[i].w;                if(!vis[v] && dist[v] > dist[u] + w) {                    dist[v] = dist[u] + w;                    Q.push(QNode(v, dist[v]));                }            }        }    }} dij[2];int main() {#ifdef ___LOCAL_WONZY___    freopen("input.txt", "r", stdin);#endif // ___LOCAL_WONZY___    int u, v, w;    while(scan_d(n) && scan_d(m) && scan_d(x)) {        dij[0].init();        dij[1].init();        for(int i = 1; i <= m; ++i) {            scan_d(u), scan_d(v), scan_d(w);            dij[0].add_edge(u, v, w);            dij[1].add_edge(v, u, w);        }        dij[0].run(x);        dij[1].run(x);        int ans = 0;        for(int i = 1; i <= n; ++i) {            if(i == x) continue;            if(dij[0].dist[i] == INF || dij[1].dist[i] == INF) continue;            ans = max(ans, dij[0].dist[i] + dij[1].dist[i]);        }        println(ans);    }    return 0;}