NightMare2（SCU4527+dijkstra+二分）

题目链接：http://acm.scu.edu.cn/soj/problem.action?id=4527

题目：

题意：最短路的每条边除了边权之外还会有一个限制（财富，身上带的财富大于这个值则不能通过这条边），问能否在k的时间内逃离迷宫，能的话最多能携带多少财富。

思路：二分最终能携带的财富值，然后跑dijkstra。

代码实现如下：

 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <bitset>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long ll;
 typedef pair<ll, ll> pll;
 typedef pair<ll, int> pli;
 typedef pair<int, ll> pil;;
 typedef pair<int, int> pii;
 typedef unsigned long long ull;

 #define lson i<<1
 #define rson i<<1|1
 #define bug printf("*********\n");
 #define FIN freopen("D://code//in.txt", "r", stdin);
 #define debug(x) cout<<"["<<x<<"]" <<endl;
 #define IO ios::sync_with_stdio(false),cin.tie(0);

 const double eps = 1e-;
 const int mod = ;
 const int mx = 1e4 + ;
 const int maxn = 5e4 + ;
 const double pi = acos(-);
 const int inf = 0x3f3f3f3f;
 const ll INF = 0x3f3f3f3f3f3f3f;

 int t, n, m, k, tot, u, v, w;
 ll cap, ans;
 int head[mx], vis[mx];
 ll dis[mx];

 struct edge {
     int v, w, next;
     ll cap;
 }ed[maxn<<];

 void addedge(int u, int v, int w, int cap) {
     ed[tot].v = v;
     ed[tot].w = w;
     ed[tot].cap = cap;
     ed[tot].next = head[u];
     head[u] = tot++;
     ed[tot].v = u;
     ed[tot].w = w;
     ed[tot].cap = cap;
     ed[tot].next = head[v];
     head[v] = tot++;
 }

 void init() {
     tot = ;
     memset(head, -, sizeof(head));
 }

 bool dij(ll x) {
     memset(dis, inf, sizeof(dis));
     memset(vis, , sizeof(vis));
     priority_queue<pli, vector<pli>, greater<pli> > q;
     q.push(make_pair(, ));
     dis[] = ;
     while(!q.empty()) {
         int u = q.top().second; q.pop();
         if(vis[u]) continue;
         vis[u] = ;
         for(int i = head[u]; ~i; i = ed[i].next) {
             int v = ed[i].v;
             if(ed[i].cap >= x && dis[v] > dis[u] + ed[i].w) {
                 dis[v] = dis[u] + ed[i].w;
                 q.push(make_pair(dis[v], v));
             }
         }
     }
     return dis[n] <= k;
 }

 int main() {
     //FIN;
     scanf("%d", &t);
     while(t--) {
         scanf("%d%d%d", &n, &m, &k);
         init();
         ll ub = , lb = , mid;
         for(int i = ; i <= m; i++) {
             scanf("%d%d%lld%d", &u, &v, &cap, &w);
             addedge(u, v, w, cap);
             if(cap > ub) ub = cap;
         }
         ans = -;
         while(ub >= lb) {
             mid = (ub + lb) >> ;
             if(dij(mid)) {
                 lb = mid + ;
                 ans = mid;
             }
             else ub = mid - ;
         }
         if(ans != -) printf("%lld\n", ans);
         else printf("Poor RunningPhoton!\n");
     }
     return ;
 }