UVA 544 - Heavy Cargo（最小瓶颈路）

UVA 544 - Heavy Cargo

题目链接

题意：就是求最小瓶颈路了。。

思路：裸题，利用kruskal性质把边从大到小加入，然后一旦连通就结束输出答案

代码：

#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;

const int N = 205;
const int M = 20005;

map<string, int> hash;
string a, b;
int hn;

int get(string a) {
if (!hash.count(a)) hash[a] = hn++;
return hash[a];
}

struct Edge {
int u, v, val;
void read() {
cin >> a >> b >> val;
u = get(a); v = get(b);
}
} e[M];

bool cmp(Edge a, Edge b) {
return a.val > b.val;
}

int n, m, parent
;

int find(int x) {
return x == parent[x] ? x : parent[x] = find(parent[x]);
}

int main() {
int cas = 0;
while (~scanf("%d%d", &n, &m) && n || m) {
string a, b, val;
hash.clear(); hn = 1;
for (int i = 1; i <= n; i++) parent[i] = i;
for (int i = 0; i < m; i++) e[i].read();
sort(e, e + m, cmp);
cin >> a >> b;
int ss = get(a), ee = get(b);
for (int i = 0; i < m; i++) {
int pu = find(e[i].u);
int pv = find(e[i].v);
if (pu != pv) {
parent[pu] = pv;
if (find(ss) == find(ee)) {
printf("Scenario #%d\n", ++cas);
printf("%d tons\n\n", e[i].val);
break;
}
}
}
}
return 0;
}