WHUST 2017 Div.2 Day 15 [2017-08-06]（图论八题）

A - Silver Cow Party（SSSP）

POJ - 3268

题目大意

有1~N一共N个结点，每个结点都有一头奶牛，

每只奶牛都打算到结点X（1<=X<=N）来参加派对，

并且在派对结束后回到原来的结点，

每只奶牛都选择最短的路程，

一共有M条单向的路，每条路通过的时间为Ti，（单向有权）

(1 ≤ N ≤ 1000)    (1 ≤ M ≤ 100,000)    (1 ≤ T_i ≤ 100)

问：来回的路程最长的奶牛经过的总路程为多少？

做法

单源最短路（SSSP）问题的变种，“回家”即为从X到各点的最短路，“前往派对”即为将所有道路取反后，从X到各个点的最短路
比较稀疏的图，用Dijkstra模板即可

AC-Code (C++)

Time: 47 ms  Memory: 1060 kb

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<queue>#include<stack>#include<string>#include<vector>#include<functional>using namespace std;const int maxn = 1005;const int maxm = 100005;const int inf = 1000000000;int N, M, X;//结点数，道路数，原点struct Edge{int from, to, dist;Edge(int u, int v, int d) :from(u), to(v), dist(d){}};vector<Edge> edges_1;//存储正边vector<int> G_1[maxn];vector<Edge> edges_2;//存储反边vector<int> G_2[maxn];bool done[maxn];//是否已被访问int d_1[maxn];//到原点的距离int d_2[maxn];typedef pair<int, int> pii;int ans = -1;void AddEdge_1(int from, int to, int dist){edges_1.push_back(Edge(from, to, dist));int m = edges_1.size();G_1[from].push_back(m - 1);}void AddEdge_2(int from, int to, int dist){edges_2.push_back(Edge(from, to, dist));int m = edges_2.size();G_2[from].push_back(m - 1);}void dijkstra_1(void)//从原点到各点{priority_queue<pii, vector<pii>, greater<pii> >Q;for (int i = 1; i <= N; i++)d_1[i] = inf;d_1[X] = 0;memset(done, 0, sizeof(done));Q.push(pii(0, X));while (!Q.empty()){pii x = Q.top();Q.pop();int u = x.second;//u为当前结点的编号if (done[u])continue;done[u] = true;for (int i = 0; i < G_1[u].size(); i++){Edge& e = edges_1[G_1[u][i]];if (d_1[e.to]>d_1[u] + e.dist){d_1[e.to] = d_1[u] + e.dist;Q.push(pii(d_1[e.to], e.to));}}}}void dijkstra_2(void)//从各点到原点{priority_queue<pii, vector<pii>, greater<pii> >Q;for (int i = 1; i <= N; i++)d_2[i] = inf;d_2[X] = 0;memset(done, 0, sizeof(done));Q.push(pii(0, X));while (!Q.empty()){pii x = Q.top();Q.pop();int u = x.second;//u为当前结点的编号if (done[u])continue;done[u] = true;for (int i = 0; i < G_2[u].size(); i++){Edge& e = edges_2[G_2[u][i]];if (d_2[e.to]>d_2[u] + e.dist){d_2[e.to] = d_2[u] + e.dist;Q.push(pii(d_2[e.to], e.to));}}}}int main(void){scanf("%d %d %d", &N, &M, &X);for (int i = 0; i < M; i++){int a, b, c;scanf("%d %d %d", &a, &b, &c);AddEdge_1(a, b, c);AddEdge_2(b, a, c);}dijkstra_1();dijkstra_2();for (int i = 1; i <= N; i++){ans = max(ans, d_1[i] + d_2[i]);}printf("%d\n", ans);return 0;}

B - Wormholes（Bellman-Ford）

POJ - 3259

题目大意

农场主FJ有F个农场（F组数据），

一个农场有N块地（结点），编号1~N，

地与地之间有M条双向正权边，W条单向负权边（虫洞），

FJ想从农场的某点出发，经过一些结点后回到原点时，时间早于出发的时刻，

问：FJ能否在该农场实现这个愿望？

(1 ≤ N ≤ 500)    (1 ≤ M ≤ 2500)    (1 ≤ W ≤ 200)

做法

要使结束的时间早于开始的时间，即为图中含有负圈，即检测给定图中是否有负圈。

Bellman-Ford模板题，

要注意输入正负边时的细节，以及总边数的处理，防止边数组开小了，另外注意一下重边的情况

AC-Code (C++)

Time: 125 ms  Memory: 764 kb

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<queue>#include<stack>#include<string>#include<vector>#include<functional>using namespace std;const int MAXN = 505;const int MAXM = 5500;const int INF = 1000000000;int F, N, M, W;//样例数，结点数，正边数（双向），负边数（单向）int d[MAXM];//到起点距离int u[MAXM];//起点int v[MAXM];//终点int w[MAXM];//权值bool ans = false;void bellman_ford(void){for (int k = 0; k <= N - 1; k++){if (k == N - 1){ans = true;break;}bool temp = false;for (int i = 1; i <= M * 2 + W; i++){int x = u[i], y = v[i];if (d[x] < INF){int dy = d[y];d[y] = min(d[y], d[x] + w[i]);if (d[y]!=dy)temp = true;}}if (!temp)break;}return;}int main(void){scanf("%d", &F);while (F--){scanf("%d %d %d", &N, &M, &W);for (int i = 1; i <= M * 2 + W; i++){w[i] = INF;}for (int i = 1; i <= N; i++){d[i] = INF;}d[1] = 0;ans = false;for (int i = 1; i <= M; i++){int a, b, c;scanf("%d %d %d", &a, &b, &c);int j = 2 * i - 1;u[j] = a;v[j] = b;w[j] = min(w[j], c);//解决重边j++;u[j] = b;v[j] = a;w[j] = min(w[j], c);}for (int i = M * 2 + 1; i <= 2 * M + W; i++){int a, b, c;scanf("%d %d %d", &a, &b, &c);u[i] = a;v[i] = b;w[i] = min(w[i], -1*c);}bellman_ford();if (ans)printf("YES\n");elseprintf("NO\n");}return 0;}

C - Cow Contest

POJ - 3660

题目大意

有1~N一共N个结点，每个结点都有一头奶牛，

每只奶牛都打算到结点X（1<=X<=N）来参加派对，

并且在派对结束后回到原来的结点，

每只奶牛都选择最短的路程，

一共有M条单向的路，每条路通过的时间为Ti，（单向有权）

(1 ≤ N ≤ 1000)    (1 ≤ M ≤ 100,000)    (1 ≤ T_i ≤ 100)

问：来回的路程最长的奶牛经过的总路程为多少？

做法

单源最短路（SSSP）问题的变种，“回家”即为从X到各点的最短路，“前往派对”即为将所有道路取反后，从X到各个点的最短路
比较稀疏的图，用Dijkstra模板即可

AC-Code (C++)

Time: 47 ms  Memory: 1060 kb