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：
1 #include <set> 2 #include <map> 3 #include <queue> 4 #include <stack> 5 #include <cmath> 6 #include <bitset> 7 #include <cstdio> 8 #include <string> 9 #include <vector> 10 #include <cstdlib> 11 #include <cstring> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 16 typedef long long ll; 17 typedef pair<ll, ll> pll; 18 typedef pair<ll, int> pli; 19 typedef pair<int, ll> pil;; 20 typedef pair<int, int> pii; 21 typedef unsigned long long ull; 22 23 #define lson i<<1 24 #define rson i<<1|1 25 #define bug printf("********* "); 26 #define FIN freopen("D://code//in.txt", "r", stdin); 27 #define debug(x) cout<<"["<<x<<"]" <<endl; 28 #define IO ios::sync_with_stdio(false),cin.tie(0); 29 30 const double eps = 1e-8; 31 const int mod = 10007; 32 const int mx = 1e4 + 7; 33 const int maxn = 5e4 + 7; 34 const double pi = acos(-1); 35 const int inf = 0x3f3f3f3f; 36 const ll INF = 0x3f3f3f3f3f3f3f; 37 38 int t, n, m, k, tot, u, v, w; 39 ll cap, ans; 40 int head[mx], vis[mx]; 41 ll dis[mx]; 42 43 struct edge { 44 int v, w, next; 45 ll cap; 46 }ed[maxn<<2]; 47 48 void addedge(int u, int v, int w, int cap) { 49 ed[tot].v = v; 50 ed[tot].w = w; 51 ed[tot].cap = cap; 52 ed[tot].next = head[u]; 53 head[u] = tot++; 54 ed[tot].v = u; 55 ed[tot].w = w; 56 ed[tot].cap = cap; 57 ed[tot].next = head[v]; 58 head[v] = tot++; 59 } 60 61 void init() { 62 tot = 0; 63 memset(head, -1, sizeof(head)); 64 } 65 66 bool dij(ll x) { 67 memset(dis, inf, sizeof(dis)); 68 memset(vis, 0, sizeof(vis)); 69 priority_queue<pli, vector<pli>, greater<pli> > q; 70 q.push(make_pair(0, 1)); 71 dis[1] = 0; 72 while(!q.empty()) { 73 int u = q.top().second; q.pop(); 74 if(vis[u]) continue; 75 vis[u] = 1; 76 for(int i = head[u]; ~i; i = ed[i].next) { 77 int v = ed[i].v; 78 if(ed[i].cap >= x && dis[v] > dis[u] + ed[i].w) { 79 dis[v] = dis[u] + ed[i].w; 80 q.push(make_pair(dis[v], v)); 81 } 82 } 83 } 84 return dis[n] <= k; 85 } 86 87 int main() { 88 //FIN; 89 scanf("%d", &t); 90 while(t--) { 91 scanf("%d%d%d", &n, &m, &k); 92 init(); 93 ll ub = 0, lb = 0, mid; 94 for(int i = 1; i <= m; i++) { 95 scanf("%d%d%lld%d", &u, &v, &cap, &w); 96 addedge(u, v, w, cap); 97 if(cap > ub) ub = cap; 98 } 99 ans = -1; 100 while(ub >= lb) { 101 mid = (ub + lb) >> 1; 102 if(dij(mid)) { 103 lb = mid + 1; 104 ans = mid; 105 } 106 else ub = mid - 1; 107 } 108 if(ans != -1) printf("%lld ", ans); 109 else printf("Poor RunningPhoton! "); 110 } 111 return 0; 112 }