POJ 1797 (最短路变形)

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题目大意：有n个城市，m条道路，在每条道路上有一个承载量，现在要求从1到n城市最大承载量，而最大承载量就是从城市1到城市n所有通路上的最大承载量

解题思路：相当于让选择的那条路最小值尽量大,更改一下最短路即可

代码:

#include <iostream>#include <cstdio>#include <cstring>#include <map>#include <set>#include <vector>#include <stack>#include <queue>#include <algorithm>#include <cmath>#define LL long long#define INF 0x3f3f3f3f#define maxn 1024#define Pair pair<int, int>using namespace std;struct edge{    int to, cost;};int V;int Test, m, test;vector<edge> G[maxn];int d[maxn];bool operator<(const Pair &a, const Pair &b){    return a.first < b.first;}void dijkstra(int s){    priority_queue<Pair> que;    fill (d, d + V + 2, 0);    d[s] = INF;    int as = INF;    que.push(Pair(INF, s));    while (!que.empty()) {        Pair p = que.top();        que.pop();        int v = p.second;        if(v == V) break;         if (d[v] > p.first) continue;         for (int i = 0; i < G[v].size(); ++i) {            edge e = G[v][i];            if(d[e.to] < min(d[v], e.cost)) {                d[e.to] = min(d[v], e.cost);                que.push(Pair(d[e.to], e.to));            }         }    }    printf("Scenario #%d:\n", test);    printf("%d\n\n", d[V]);}int main(){    scanf("%d", &Test);    test = 0;    while (test != Test) {        test++;        scanf("%d%d", &V, &m);        for(int i = 0; i <= V; ++i)        G[i].clear();        int from, to, cost;        for (int i = 1; i <= m; ++i) {            scanf("%d%d%d", &from, &to, &cost);            edge demo;            demo.to = to;            demo.cost = cost;            G[from].push_back(demo);            demo.to = from;            G[to].push_back(demo);        }        dijkstra(1);    }    return 0;}

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