這題要處理的地方的就是一個(gè)人可以同時(shí)向多個(gè)人傳遞消息，也就是說一條消息的傳遞時(shí)間由最長(zhǎng)的那一條路徑所決定，因?yàn)榭梢酝瑫r(shí)進(jìn)行嘛，所以就求某一點(diǎn)到所有點(diǎn)的最短路，然后再尋找一條最長(zhǎng)的路勁，枚舉每個(gè)頂點(diǎn)作為起點(diǎn)就可以了。

代碼如下：

#include<iostream> #include<cstdio> #include<cstdlib> #include<algorithm> #include<cmath> #include<queue> #include<set> #include<map> #include<cstring> #include<vector> #include<string> #define inf 0x3f3f3f3f #define LL long long using namespace std;int N, G[105][105], dis[105], vis[105];bool Dijkstra(int s, int &ret) {int pos, Min;ret = 0;memset(dis, 0x3f, sizeof (dis));memset(vis, 0, sizeof (vis));dis[s] = 0;for (int i = 1; i <= N; ++i) {Min = inf;for (int j = 1; j <= N; ++j) {if (!vis[j] && Min > dis[j]) {pos = j, Min = dis[j];}}if (Min == inf) {return false;} else {vis[pos] = 1;for (int j = 1; j <= N; ++j) {if (!vis[j] && G[pos][j] != inf && dis[j] > dis[pos] + G[pos][j]) {dis[j] = dis[pos] + G[pos][j]; }}}}for (int i = 1; i <= N; ++i) {ret = max(ret, dis[i]); }return true; }int main() {int a, b, M, flag, Min, num, ret;while (scanf("%d", &N), N) {flag = 0;Min = inf;memset(G, 0x3f, sizeof (G));for (int i = 1; i <= N; ++i) {scanf("%d", &M);for (int j = 1; j <= M; ++j) {scanf("%d %d", &a, &b);G[i][a] = b;}} for (int i = 1; i <= N; ++i) {int t;if (Dijkstra(i, t)) {flag = 1;if (Min > t) {Min = t, num = i;}}}if (!flag) puts("disjoint");else printf("%d %d\n", num, Min);}return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/Lyush/archive/2012/09/27/2705242.html

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