学步园

题意 ： 给你一张有向图， 然后在给你k条新的路和一个d，问你在建立不超过d条新路条件下节点0 到节点n - 1 的最短路。

思路 ： 这道题目可以转换成有限制的最短路来做的，只要把新边认为w = 1,而旧边就是w = 0,最后w总和不超过 d后就是裸裸的有限制最短路，至于有限制的最短路可以用A* 优化做的，评估函数 是f(x) = g(x) + h(x) 其中g(x)表示的是当前搜索到的满足条件的路径长度，而h(x)表示 x 到节点n - 1的最短路径用Dijkstra建立反向边预处理出来，评估函数满足评估值小于等于实际值这一条件。

#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int INF = 83178122;
const int maxn = 100005;

struct AS{
    int g, h, x, num;
    AS(){}
    AS(int a, int b, int c, int d) : g(a), h(b), x(c), num(d){}
    bool operator < (const AS &cmp)const {
        return g + h > cmp.g + cmp.h;
    }
};

struct Dist{
    int d, x;
    Dist(){}
    Dist(int a, int b) : d(a), x(b){}
    bool operator < (const Dist &cmp)const {
        return d > cmp.d;
    }
};

struct Edge{
    int to, c, w, next;
}edge[2][maxn<<1];

int head[2][maxn], E[2];
int n, m, k, Lim;
int vis[maxn], d[maxn];

void init(){
    E[0] = E[1] = 0;
    memset(head, -1, sizeof(head));
}

void add_edge(int op, int u, int to, int c, int w){
    edge[op][E[op]].to = to; edge[op][E[op]].w = w;
    edge[op][E[op]].c = c; edge[op][E[op]].next = head[op][u];
    head[op][u] = E[op]++;
}

void Dijkstra(){
    fill(d, d+n, INF); d[n-1] = 0;
    memset(vis, 0, sizeof(vis));
    priority_queue<Dist> q;
    q.push(Dist(d[n-1], n-1));
    while (!q.empty()){
        Dist tmp = q.top(); q.pop();
        int u = tmp.x;
        if (vis[u])continue;
        vis[u] = 1;
        for (int i = head[1][u]; i != -1;  i = edge[1][i].next){
            int to = edge[1][i].to, c = edge[1][i].c;
            if (d[to] > d[u] + c){
                d[to] = d[u] + c;
                q.push(Dist(d[to], to));
            }
        }
    }
    return ;
}

int A_star(){
    Dijkstra();
    priority_queue<AS> q;
    q.push(AS(0, d[0], 0, 0));
    int ret = INF;
    while (!q.empty()){
        AS tmp = q.top(); q.pop();
        int u = tmp.x;
        if (tmp.num > Lim)continue;
        if (u == n - 1)return tmp.g;
        for (int i = head[0][u]; i != -1; i = edge[0][i].next){
            int to = edge[0][i].to, c = edge[0][i].c, w = edge[0][i].w;
            q.push(AS(tmp.g+c, d[to], to, tmp.num+w));
        }
    }
    return ret;
}

int main(){
    int T;
    scanf("%d", &T);
    for (int cas = 1; cas<= T; cas++){
        scanf("%d%d%d%d", &n, &m, &k, &Lim);
        init();
        for (int i = 1; i <= m; i++){
            int a, b, c;
            scanf("%d%d%d", &a, &b, &c);
            add_edge(0, a, b, c, 0);
            add_edge(1, b, a, c, 0);
        }
        for (int i = 1; i <= k; i++){
            int a, b, c;
            scanf("%d%d%d", &a, &b, &c);
            add_edge(0, a, b, c, 1);
            add_edge(1, b, a, c, 1);
        }
        int ret = A_star();
        if (ret >= INF)printf("Case %d: Impossible\n", cas);
        else printf("Case %d: %d\n", cas, ret);
    }
    return 0;
}