【POJ】1502 MPI Maelstrom

题目大意就是找到第一个结点到其他结点最短距离的最大值，所以首先使用Dijkstra算法找到第一个结点到其他结点的最短距离，之后再找到最大值即可：

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <queue>

using namespace std;

const int N = 1000 + 10;
const int INF = 999999;

struct Edge{
    
    int to, length;
    Edge(int t, int l): to(t), length(l){
    }
};

struct Point{
    
    int number;
    int distance;
    Point(int n, int d): number(n), distance(d){
    }
    bool operator<(const Point &p) const{
    
        return distance > p.distance;
    }
};
vector<Edge> graph[N];
int dis[N];

void Dijkstra(int s){
    
    priority_queue<Point> pq;
    dis[s] = 0;
    pq.push(Point(s, dis[s]));
    while(!pq.empty()){
    
        int u = pq.top().number;
        pq.pop();
        for(int i = 0; i < graph[u].size(); i ++ ){
    
            int v = graph[u][i].to;
            int l = graph[u][i].length;
            if(dis[v] > dis[u] + l){
    
                dis[v] = dis[u] + l;
                pq.push(Point(v, dis[v]));
            }
        }
    }
}

int Toint(string s){
    
    int n = 0;
    for(int i = 0; i < s.size(); i ++ ){
    
        n = n * 10 + s[i] - '0';
    }
    return n;
}

int main(){
    
    int n;
    while(cin >> n){
    
        string ch;
        for(int i = 2; i <= n; i ++ ){
    
            for(int j = 1; j < i; j ++ ){
    
                cin >> ch;
                if(ch != "x"){
    
                    graph[i].push_back(Edge(j, Toint(ch)));
                    graph[j].push_back(Edge(i, Toint(ch)));
                }
            }
        }
        memset(dis, INF, sizeof(dis));
        Dijkstra(1);
        int maximum = -1;
        for(int i = 2; i <= n; i ++ ) maximum = max(maximum, dis[i]);
        cout << maximum << endl;
    }
    return 0;
}