poj3169 Layout

/*  这次去查了一下，为什么大家要那样设置INF，有了些新发现  http://blog.csdn.net/jiange_zh/article/details/50198097    以及，这题的思路来自挑战，这个思路真是极其巧妙，将奶牛之间的最短距离、最长距离都转换为了最短路问题*/#include <iostream>#include <cstdio>const int MAX_ML = 1e4 + 10;const int MAX_MD = 1e4 + 10;const int INF = 0x3f3f3f3f;const int MAX_N = 1010;int N, ML, MD;int AL[MAX_ML], BL[MAX_ML], DL[MAX_ML];int AD[MAX_MD], BD[MAX_MD], DD[MAX_MD];int d[MAX_N]; //最短距离 using namespace std;void solve(){fill(d, d + N, INF);d[0] = 0;// 用Bellman-Ford算法计算dfor (int k = 0; k < N; k++){// 从 i+1 到 i 的权值为0for (int i = 0; i + 1 < N; i++)if (d[i + 1] < INF) d[i] = min(d[i], d[i + 1]);// 从 AL 到 BL 的权值为 DLfor (int i = 0; i < ML; i++){if (d[AL[i] - 1] < INF)d[BL[i] - 1] = min( d[BL[i] - 1], d[AL[i] - 1] + DL[i] );}// 从 BD 到 AD 的权值为 -DD for (int i = 0; i < MD; i++){if (d[BD[i] - 1] < INF)d[AD[i] - 1] = min ( d[AD[i] - 1], d[BD[i] - 1] - DD[i] );}}int res = d[N - 1];if (d[0] < 0) //存在负圈无解cout << -1 << endl;else if (res == INF)cout << -2 << endl;else cout << res << endl; }int main(){cin >> N >> ML >> MD;for (int i = 0; i < ML; i++)cin >> AL[i] >> BL[i] >> DL[i];for (int i = 0; i < MD; i++)cin >> AD[i] >> BD[i] >> DD[i];solve();return 0;}