POJ 3268 Silver Cow Party

Description

One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; roadi requires T_i (1 ≤T_i ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input

Line 1: Three space-separated integers, respectively:N, M, and X
Lines 2.. M+1: Line i+1 describes road i with three space-separated integers:A_i, B_i, andT_i. The described road runs from farmA_i to farm B_i, requiringT_i time units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

Sample Input

4 8 21 2 41 3 21 4 72 1 12 3 53 1 23 4 44 2 3

Sample Output

10

Hint

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

题目大意：

在一张有向图中n - 1头牛去另外一头牛家参加派对， 并会在结束后返回自己的家中， 但是他们都很懒， 所以会走效率高的路（尽可能短的路）， 请求出走得最远的牛走了多远。

解题思路：

一开始用Dijkstra傻傻的从单源点向其他点跑和从其他点向单源点跑， 妥妥TLE， 按点跑Dijkstra的时间复杂度是O（N^2）。 于是我们可以开两个数组， 一个记录所有牛回家的距离， 一个记录所有牛走的距离。 两个距离相加取最大值， 对于从其他点向单源点跑的操作， 我们可以对于初始矩阵进行矩阵转置操作， 相当于在反着跑一边。O（2 * N）。第二种解法可以用优先队列， 按着边开始跑，时间复杂度为O（N * (Elog(E))）。

代码：

#include <iostream>
#include <sstream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iomanip>
#include <utility>
#include <string>
#include <cmath>
#include <vector>
#include <bitset>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>

using namespace std;

/*
 *ios::sync_with_stdio(false);
 */

typedef long long ll;
typedef unsigned long long ull;
const int inf = 0x7fffffff;
const int mod = 1000000;
const int Max = 1010;

int n, m ,x;
int Map[Max][Max][2];
bool vis[Max];
int dis_c[Max], dis_g[Max];

void Init() {
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            Map[i][j][0] = (i == j) ? 0 : inf;
            //Map[j][i][0] = (i == j) ? 0 : inf;
            Map[i][j][1] = (i == j) ? 0 : inf;
            //Map[j][i][1] = (i == j) ? 0 : inf;
        }
    }
}

void map_Swap() {
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            Map[i][j][0] = Map[j][i][1];
        }
    }
}

void Dijkstra(int st, int f) {
    int dis[Max];
    for (int i = 1; i <= n; ++i) {
        dis[i] = Map[st][i][f];
        vis[i] = 0;
    }
    vis[st] = 1;
    for (int i = 1; i <= n; ++i) {
        int Min = inf, mark;
        for (int j = 1; j <= n; ++j) {
            if (!vis[j] && Min > dis[j]) {
                Min = dis[mark = j];
            }
        }
        vis[mark] = true;
        for (int j = 1; j <= n; ++j) {
                        // that's very important
            if (!vis[j] && Map[mark][j][f] != inf && dis[j] > dis[mark] + Map[mark][j][f]) {
                dis[j] = dis[mark] + Map[mark][j][f];
            }
        }
    }
    if (f == 0) {
        for (int i = 1; i <= n; ++i) {
            dis_c[i] = dis[i];
        }
    }
    else{
        for (int i = 1; i <= n; ++i) {
            dis_g[i] = dis[i];
        }
    }
}

void check() {
    int ans = -1;
    for (int i = 1; i <= n; ++i) {
        ans = max(ans, (dis_c[i] + dis_g[i]));
    }
    printf("%d\n", ans);
}

int main() {
    // initialization
    scanf("%d %d %d", &n, &m, &x);
    Init();
    for (int i = 0; i < m; ++i) {
        int x, y, t;
        scanf("%d %d %d", &x, &y, &t);
        if (Map[x][y][1] > t)
            Map[x][y][1] = t;
    }
    // back
    Dijkstra(x, 1);
    // a to b >> b to a
    map_Swap();
    // go
    Dijkstra(x, 0);
    // output the answer
    check();
    return 0;
}

/** Dijkstra + priority_queue

#include <iostream>
#include <sstream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iomanip>
#include <utility>
#include <string>
#include <cmath>
#include <vector>
#include <bitset>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>

using namespace std;

/*
 *ios::sync_with_stdio(false);
 */

typedef long long ll;
typedef unsigned long long ull;
const int inf = 0x7fffffff;
const int mod = 1000000;
const int Max = 1010;

int n, m ,x;
int dis[Max];
bool vis[Max];
struct node {
    // endpoint and distance
    int x, d;
    node() {};
    node(int a, int b) {
        x = a;
        d = b;
    }
    bool operator < (const node &a) const {
        if (d == a.d) return x < a.x;
        else return d > a.d;
    }
};

vector < node> edge[Max];

void Init() {
    for (int i = 0; i <= n; ++i) {
        edge[i].clear();
    }
}

int Dijkstra(int st, int ed) {
    for (int i = 0; i <= n; ++i) {
       dis[i] = inf;
    }
    dis[st] = 0;
    priority_queue < node> Q;
    Q.push(node(st, dis[st]));
    while (!Q.empty()) {
        node x = Q.top();
        Q.pop();
        for (int i = 0; i < edge[x.x].size(); ++i) {
            node y = edge[x.x][i];
            if (dis[y.x] > x.d + y.d) {
                dis[y.x] = x.d + y.d;
                Q.push(node(y.x, dis[y.x]));
            }
        }
    }
    return dis[ed];
}

void check() {
    int ans = 0, minn;
    for (int i = 1; i <= n; ++i) {
        minn = Dijkstra(x, i) + Dijkstra(i, x);
        if (minn > ans)
            ans = minn;
    }
    printf("%d\n", ans);
}

int main() {
    // initialization
    scanf("%d %d %d", &n, &m, &x);
    Init();

    for (int i = 0; i < m; ++i) {
        int x, y, t;
        scanf("%d %d %d", &x, &y, &t);
        edge[x].push_back(node(y, t));
    }
    // perform the algorithm
    check();
    return 0;
}
*/