NYOJ-183 赚钱啦【最短路】

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题目链接：http://acm.nyist.net/JudgeOnline/problem.php?pid=183

解题思路：

最短路裸题。

最长路+判断正环

学会的东西是如何用SPFA判断负环，可以用一个数组维护每个节点入队的次数，超过n次就肯定有负环出现。

代码如下：

#include<iostream>#include<cstdio>#include<cmath>#include<queue>#include<climits>#include<cstring>#include<string>#include<algorithm>using namespace std;const int MAX = INT_MAX >> 1;const int N = 1010;const int M = 2010;#define CLR(arr, what) memset(arr, what, sizeof(arr))int Key[M], Next[M], Num[M], Head[N], top;int dis[N], visit[N], relaxnum[N];int n, m;void init(){CLR(Head, -1);CLR(Next, -1);CLR(relaxnum, 0);top = 0;}void add(int u, int v, int cost){Key[top] = cost;Next[top] = Head[u];Num[top] = v;Head[u] = top++;}int SPFA(int start){queue<int> q;while(!q.empty())q.pop();CLR(visit, false);for(int i = 0; i < N; ++i)dis[i] = -MAX;dis[start] = 0;visit[start] = true;q.push(start);while(!q.empty()){int cur = q.front();q.pop();visit[cur] = false;for(int i = Head[cur]; i != -1; i = Next[i]){if(dis[Num[i]] < dis[cur] + Key[i]){dis[Num[i]] = dis[cur] + Key[i];if(!visit[Num[i]]){q.push(Num[i]);if(++relaxnum[Num[i]] > n) //正无穷环return 1;visit[Num[i]] = true;}}}}return 0;}int main(){int ncase;bool flag;int start, end, spend, to, back;scanf("%d", &ncase);while(ncase--){init();flag = false;scanf("%d%d", &n, &m);for(int i = 0; i < m; ++i){scanf("%d%d%d%d%d", &start, &end, &spend, &to, &back);if((to - spend > 0 && back - spend >= 0) || (to - spend >= 0 && back - spend > 0) || (to - spend > 0 && back - spend > 0)){flag = true;continue;}add(start, end, to - spend);add(end, start, back - spend);}if(flag == true || SPFA(0))printf("$$$\n");elseprintf("%d\n", dis[n - 1]);}return 0;}