[POJ 1724]ROADS[SPFA][DFS剪枝]

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题目链接：[POJ 1724]ROADS[SPFA][DFS剪枝]

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;}

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