[poj] 1797

Heavy Transportation

Time Limit: 3000MS Memory Limit: 30000KTotal Submissions: 40609 Accepted: 10661

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 output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the maximum allowed weight that Hugo can transport to the customer. Terminate the output for the scenario with a blank line.

Sample Input

13 31 2 31 3 42 3 5

Sample Output

Scenario #1:4

给你一个图，任意两个节点直接有桥梁相连，每个桥梁有一个最大承重，问你若要从 1 号 到 n 号卡车的最大载重是多少。

最早接触这题是朗讯杯

当时是二分答案 + BFS

然后之前做的一题给了我灵感

二份答案 + 并查集

先把所有边排序，然后二分

判断是否可达

刚开始从小到大排序 WA 飞了

然后这题 m 的范围没给

RE 飞了

体验极差

#include <cstdio>#include <algorithm>#include <cstring>#include <vector>#include <queue>#include <iostream>#define pb push_back#define mp make_pair#define LL long longusing namespace std;///重边  自环const int N = 100010;struct Node{    int u;    int v;    int w;};bool cmp(Node a, Node b){    return a.w > b.w;}int n, m;int ne[N];Node node[N];void ini(){    for(int i = 1; i <= n; i ++){        ne[i] = i;    }}int Find(int x){    int t = x;    while(t != ne[t]){        t = ne[t];    }    while(t != x){        int q;        q = ne[x];        ne[x] = t;        x = q;    }    return t;}void Join(int x, int y){    x = Find(x);    y = Find(y);    if(x != y){        ne[x] = y;    }}int main(){    int ncase;    int nc = 1;    int ans;    int l;    int r;    scanf("%d", &ncase);    while(ncase --){        scanf("%d%d", &n, &m);        for(int i = 0; i < m; i ++){            scanf("%d%d%d", &node[i].u, &node[i].v, &node[i].w);        }        sort(node, node + m, cmp);        l = 0;        r = m - 1;        while(l <= r){            int mid = l + ((r - l) >> 1);            ini();            for(int i = 0; i <= mid; i ++){                Join(node[i].u, node[i].v);            }            if(Find(1) == Find(n)){                ans = mid;                r = mid - 1;            }            else{                l = mid + 1;            }        }        printf("Scenario #%d:\n", nc ++);        printf("%d\n\n", node[ans].w);    }    return 0;}

P.S.

上了读入挂以后达到了 79MS

眉开眼笑