POJ 1797 Heavy Transportation (最大生成树)

标签：man   ast   for   www   follow   limit   any   tps   bridge   

题目链接：POJ 1797

Description

Background

Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has build his giant steel crane to the place where it is needed on which all streets can carry the weight.

Fortunately he already has a plan of the city with all streets and bridges and all the allowed weights.Unfortunately he has no idea how to find the the maximum weight capacity in order to tell his customer how heavy the crane may become. But you surely know.

Problem

You are given the plan of the city, described by the streets (with weight limits) between the crossings, which are numbered from 1 to n. Your task is to find the maximum weight that can be transported from crossing 1 (Hugo‘s place) to crossing n (the customer‘s place). You may assume that there is at least one path. All streets can be travelled in both directions.

Input

The first line contains the number of scenarios (city plans). For each city the number n of street crossings (1 <= n <= 1000) and number m of streets are given on the first line. The following m lines contain triples of integers specifying start and end crossing of the street and the maximum allowed weight, which is positive and not larger than 1000000. There will be at most one street between each pair of crossings.

Output

The first line contains the number of scenarios (city plans). For each city the number n of street crossings (1 <= n <= 1000) and number m of streets are given on the first line. The following m lines contain triples of integers specifying start and end crossing of the street and the maximum allowed weight, which is positive and not larger than 1000000. There will be at most one street between each pair of crossings.

Sample Input

1
3 3
1 2 3
1 3 4
2 3 5

Sample Output

Scenario #1:
4

Source

TUD Programming Contest 2004, Darmstadt, Germany

Solution

题意

有 N 个城市，M 条道路，Hugo Heavy 要从城市 1 到城市 N 运输货物，每条道路都有它的最大载重量，求从城市 1 到城市 N 运送最多的重量是多少。

思路

最大生成树

题目要求点 1 到 N 的所有路径的所有边权的最小值中的最大值。维护一个最大生成树，不断将最大权值的边加入，如果遇到点 N 就结束。

这题有点坑，输出两个换行。

此题还可以用 \(Dijkstra\) 解决。戳这里

Code

Prim

#include <iostream>
#include <cstdio>
#include <queue>
#include <map>
#include <cmath>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 1010;
const int inf = 0x3f3f3f3f;
typedef pair<int, int> P;
int n, m;
int ans;

struct Edge {
    int to, w;
    Edge(int to, int w): to(to), w(w) {}
};
vector<Edge> G[N];
int d[N], v[N];

void init() {
    for(int i = 0; i < N; ++i) {
        G[i].clear();
    }
    ans = inf;
}

void add(int x, int y, int z) {
    G[x].push_back(Edge(y, z));
}

void prim(int s) {
    priority_queue<P> q;
    memset(d, 0, sizeof(d));
    memset(v, 0, sizeof(v));
    d[s] = inf;
    q.push(P(inf, s));
    while(q.size()) {
        P p = q.top(); q.pop();
        int x = p.second;
        if(v[x]) continue;
        v[x] = 1;
        ans = min(ans, p.first);
        if(x == n) return;
        for(int i = 0; i < G[x].size(); ++i) {
            Edge e = G[x][i];
            if (d[e.to] < e.w && !v[e.to]) {
                d[e.to] = e.w;
                q.push(P(d[e.to], e.to));
            }
        }
    }
}

int main() {
    int T;
    scanf("%d", &T);
    int kase = 0;
    while(T--) {
        init();
        scanf("%d%d", &n, &m);
        for(int i = 0; i < m; ++i) {
            int x, y, z;
            scanf("%d%d%d", &x, & y, &z);
            add(x, y, z);
            add(y, x, z);
        }
        prim(1);
        if(kase) printf("\n");
        printf("Scenario #%d:\n", ++kase);
        printf("%d\n", ans);
    }
    return 0;
}

Kruskal

#include <iostream>
#include <cstdio>
#include <queue>
#include <map>
#include <cmath>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 1010, M = 1e6 + 10;
const int inf = 0x3f3f3f3f;
int n, m;
int ans;

struct Edge {
    int x, y, z;
} edge[M];

int fa[N];

int cmp(Edge a, Edge b) {
    return a.z > b.z;
}

int get(int x) {
    if(x == fa[x]) return x;
    return fa[x] = get(fa[x]);
}

void init() {
    for(int i = 0; i <= n; ++i) {
        fa[i] = i;
    }
    ans = inf;
}

void kruskal() {
    sort(edge + 1, edge + 1 + m, cmp);
    for(int i = 1; i <= m; ++i) {
        int x = get(edge[i].x);
        int y = get(edge[i].y);
        if(x != y) {
            ans = min(ans, edge[i].z);
            fa[x] = y;
        }
        x = get(1);
        y = get(n);
        if(x == y) return;
    }
}

int main() {
    int T;
    scanf("%d", &T);
    int kase = 0;
    while(T--) {
        scanf("%d%d", &n, &m);
        init();
        for(int i = 1; i <= m; ++i) {
            scanf("%d%d%d", &edge[i].x, &edge[i].y, &edge[i].z);
        }
        kruskal();
        if(kase) printf("\n");
        printf("Scenario #%d:\n", ++kase);
        printf("%d\n", ans);
    }
    return 0;
}

POJ 1797 Heavy Transportation (最大生成树)

标签：man   ast   for   www   follow   limit   any   tps   bridge   

原文地址：https://www.cnblogs.com/wulitaotao/p/11664376.html