uva 11280(Dijkstra+递推)

题意：在一张有重边的有向图中有若干城市，有若干个询问每次告诉你最多能经过q个城市，让你求最多经过q个城市从1号城市到n号城市的最短路是什么？

思路：很容易想到用Dijkstra稍作变形就能解决这一题，只要在我们计算最短路的dis数组中再增加一维，表示最多经过的城市数，这样每次放进队列中的第二个状态变成两个，最后再遍历一下dis数组求一下最小值就能得出答案了。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;

typedef long long ll;
typedef pair<int ,int> pii;
typedef pair<unsigned int, unsigned int> puu;
typedef pair<int ,double> pid;
typedef pair<ll, int> pli;

const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const int LEN = 110;
struct V{char name[22];};
typedef struct{int v, n;}St;
typedef pair<int, St> pis;
V vex[LEN];
int n, m, dis[LEN][LEN];
vector<pii> Map[LEN];

struct cmp{
    bool operator ()(pis a, pis b){return a.first>b.first;}
};

void Dijkstra(int s)
{
    priority_queue<pis, vector<pis>, cmp> q;
    int vis[LEN][LEN] = {0};St temp;
    memset(dis, 0x3f, sizeof dis);
    dis[s][0] = 0;
    temp.v = s;temp.n = 0;
    q.push(MP(dis[s][0], temp));
    while(!q.empty()){
        pis nvex = q.top(); q.pop();
        St nv = nvex.second;
        if(vis[nv.v][nv.n])continue;
        vis[nv.v][nv.n] = 1;
        for(int i=0; i<Map[nv.v].size(); i++){
            int x = Map[nv.v][i].first, y = Map[nv.v][i].second;
            if(dis[x][nv.n+1]>dis[nv.v][nv.n]+y){
                dis[x][nv.n+1] = dis[nv.v][nv.n]+y;
                temp.v = x;temp.n = nv.n+1;
                q.push(MP(dis[x][nv.n+1], temp));
            }
        }
    }
}

int main()
{
//    freopen("in.txt", "r", stdin);

    int T, val, a, b, q, qn;
    char from[22], to[22];
    scanf("%d", &T);
    for(int kase = 1; kase<=T; kase++){
        for(int i=0; i<LEN; i++)Map[i].clear();
        scanf("%d", &n);
        for(int i=1; i<=n; i++){
            scanf("%s", vex[i].name);
        }
        scanf("%d", &m);
        for(int i=0; i<m; i++){
            scanf("%s%s%d", from, to, &val);
            for(int j=1; j<=n; j++){
                if(!strcmp(vex[j].name, from))a = j;
                if(!strcmp(vex[j].name, to))b = j;
            }
            Map[a].PB(MP(b, val));
        }
        Dijkstra(1);
        printf("Scenario #%d
", kase);
        scanf("%d", &qn);
        for(int i=0; i<qn; i++){
            scanf("%d", &q);
            int ans = INF;
            for(int j=0; j<=q+1; j++){
                ans = min(ans, dis[n][j]);
            }
            if(ans!=INF)printf("Total cost of flight(s) is $%d
", ans);
            else printf("No satisfactory flights
");
        }
        if(kase!=T)printf("
");
    }
    return 0;
}