SCU2016-01 O题dijkstra变形或者最小生成树变形

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Analyse:
把边从大到小排序，然后依次加边，直到1和n连通的最小边长即可。
另一种思路：
考虑dijkstra求最长路我们知道，这里求最大承重量也行。
定义d[i]为从1到i的承重量，然后松弛操作是d[j]=min(d[i],cost(i,j))
可以知道dijkstra的性质不受松弛操作的影响而影响，因为在路径长度中松弛操作通常会把值变大，而这里承重量是变小。
可以说dijkstra只是提供了一种更新顺序，总是用最优的值去更新。

/**********************jibancanyang************************** *Author*        :jibancanyang *Created Time*  : 一  7/ 4 15:09:42 2016**Problem**:**Get**:**Code**:***********************1599664856@qq.com**********************/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <stack>using namespace std;typedef pair<int, int> pii;typedef long long ll;typedef unsigned long long ull;typedef vector<int> vi;#define pr(x) cout << #x << ": " << x << "  " #define pl(x) cout << #x << ": " << x << endl;#define pri(a) printf("%d\n",(a))#define xx first#define yy second#define sa(n) scanf("%d", &(n))#define sal(n) scanf("%lld", &(n))#define sai(n) scanf("%I64d", &(n))#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++) const int mod = int(1e9) + 7, INF = 0x3f3f3f3f;const int maxn = 1000 + 13;int par[maxn], rnk[maxn];struct edge {    int from, to, cost;};void init(int n) {    for (int i = 1; i <= n; i++) {        par[i] = i;        rnk[i] = 0;    }}int find(int x) {    if (x == par[x]) return x;    return find(par[x]);}void unite(int x, int y) {    x = find(x), y = find(y);    if (x == y) return;    if (rnk[x] < rnk[y]) {        par[x] = y;    } else {        par[y] = x;        if (rnk[x] == rnk[y]) rnk[x]++;    }}bool same(int a, int b) {    return find(a) == find(b);}bool cmp(const edge &a ,const edge &b) { return a.cost > b.cost; }int main(void){#ifdef LOCAL    freopen("in.txt", "r", stdin);    //freopen("out.txt", "w", stdout);#endif    int T; sa(T);    for (int cas = 1; cas <= T; cas++) {        int n, m;        sa(n), sa(m);        vector<edge> v;        init(n);        for (int i = 0; i < m; i++) {            edge temp;            sa(temp.from), sa(temp.to), sa(temp.cost);            v.push_back(temp);        }        sort(v.begin(), v.end(), cmp);        int ans = 0;        for (int i = 0; i < (int)v.size(); i++) {            unite(v[i].from, v[i].to);            if (same(1, n)) {                ans = v[i].cost;                break;            }        }        printf("Scenario #%d:\n", cas);        printf("%d\n\n", ans);    }    return 0;}

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