题意分析:
求从点1到点N费用不超过K的最短路。
解题思路:
用spfa根据最小花费,跑出一组距离作为起始答案,如果该答案费用大于K,那么就无解。
确认有解后,我们从点1开始DFS,用花费和之前算的距离作为剪枝。
个人感受:
差点弃疗了。1A也是蛮爽的。
具体代码如下:
#include<algorithm> #include<cctype> #include<cmath> #include<cstdio> #include<cstring> #include<iomanip> #include<iostream> #include<map> #include<queue> #include<set> #include<sstream> #include<stack> #include<string> #define ll long long #define pr(x) cout << #x << " = " << (x) << '\n'; using namespace std; const int INF = 0x3f3f3f3f; const int MAXN = 1e4 + 111; struct Edge { int to, next, l, t; }edge[MAXN]; int head[111], tol = 0, n, k, ans, cost[111], dis[111]; bool in[111]; void add_edge(int u, int v, int l, int t) { edge[tol].to = v; edge[tol].l = l; edge[tol].t = t; edge[tol].next = head[u]; head[u] = tol++; } bool spfa() { memset(in, 0, sizeof in); memset(cost, 0x3f, sizeof cost); cost[1] = 0; dis[1] = 0; in[1] = 1; queue<int> q; q.push(1); while (q.size()) { int u = q.front(); q.pop(); in[u] = 0; for (int i = head[u]; ~i; i = edge[i].next) { int v = edge[i].to; if (cost[v] > cost[u] + edge[i].t) { cost[v] = cost[u] + edge[i].t; dis[v] = dis[u] + edge[i].l; if (!in[v]) { in[v] = 1; q.push(v); } } } } return (cost[n] <= k); } void dfs(int x, int len, int cos) { if (cos > k || len >= ans) return; if (x == n) { ans = len; return; } for (int i = head[x]; ~i; i = edge[i].next) { int v = edge[i].to; dfs(v, len + edge[i].l, cos + edge[i].t); } } int main() { int r, u, v, l, t; scanf("%d%d%d", &k, &n, &r); memset(head, -1, sizeof head); for (int i = 0; i < r; ++i) { scanf("%d%d%d%d", &u, &v, &l, &t); add_edge(u, v, l, t); } if (!spfa()) printf("-1\n"); else { ans = dis[n]; dfs(1, 0, 0); printf("%d\n", ans); } return 0; }