lightoj 1254 - Prison Break

来源：互联网发布：德邦统计分析员知乎编辑：程序博客网时间：2024/06/08 05:02

题目链接：http://lightoj.com/volume_showproblem.php?problem=1254

分析题意：说的是个凡人的逃跑，有辆汽车，其邮箱容量是cap<=100。然后有n个城市，m条城市之间的距离，汽车的耗油量是1:1的。开始油箱里面是没有油的，有Q个询问，从u到v油箱的容量为cap的时候的最小花费，不能到达就输出impossible，否则输出最小花费。

这个就是多状态的最短路，据基础题。

设定状态dis[u][r]表示到达u点汽车剩余油量为r时的最小费用。在就是对于每个城市要存的数据的，剩余油量，及对应的花费，然后就是点的标号。(i,j,k)，优先队列中以费用小的优先。

typedef<pair<int,int>, int> iii;

/*****************************************Author      :Crazy_AC(JamesQi)Time        :2015File Name   :*****************************************/// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <algorithm>#include <iomanip>#include <sstream>#include <string>#include <stack>#include <queue>#include <deque>#include <vector>#include <map>#include <set>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <climits>using namespace std;#define MEM(x,y) memset(x, y,sizeof x)#define pk push_backtypedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> ii;typedef pair<ii,int> iii;const double eps = 1e-10;const int inf = 1 << 30;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int maxn =1e3 + 10;vector<ii> G[maxn];int n,m;int cap;int dis[maxn][110];int cost[maxn];int solve(int st,int ed){    priority_queue<iii, vector<iii>, greater<iii> > que;    memset(dis, 0x7f, sizeof dis);    dis[st][0] = 0;    int ret = 0;    que.push(iii(ii(dis[st][0], 0), st));    while(!que.empty()){        iii now = que.top();        que.pop();        int u = now.second;        int w = now.first.first;        int r = now.first.second;        if (w > dis[u][r]) continue;        if (u == ed) return dis[u][r];        int size = G[u].size();        for (int i = 0;i < size;++i){            int v = G[u][i].first;            int len = G[u][i].second;            if (len > cap) continue;            if (r >= len && dis[v][r - len] > dis[u][r]){                dis[v][r - len] = dis[u][r];                que.push(iii(ii(dis[v][r - len], r - len), v));            }        }        if (r < cap){            dis[u][r + 1] = dis[u][r] + cost[u];            que.push(iii(ii(dis[u][r + 1], r + 1), u));        }    }    return INF;}int main(){        // freopen("in.txt","r",stdin);    // freopen("out.txt","w",stdout);    int t, icase = 0;    cin >> t;    while(t--){        cin >> n >> m;        for (int i = 0;i < n;++i)            cin >> cost[i];        int u, v, w;        for (int i = 0;i < n;++i)            G[i].clear();        for (int i = 0;i < m;++i){            cin >> u >> v >> w;            G[u].push_back(ii(v, w));            G[v].push_back(ii(u, w));        }        printf("Case %d:\n", ++icase);        int Q;        cin >> Q;        // cout << "Q = " << Q << endl;        while(Q--){            cin >> cap >> u >> v;            int ans = solve(u, v);            if (ans == INF) puts("impossible");            else cout << ans << endl;        }    }    return 0;}

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