POJ

           思路：直接Dijkstra，求源点1到其他各点的最短距离的最大值。

AC代码

#include <cstdio>#include <cmath>#include <cctype>#include <algorithm>#include <cstring>#include <utility>#include <string>#include <iostream>#include <map>#include <set>#include <vector>#include <queue>#include <stack>using namespace std;#pragma comment(linker, "/STACK:1024000000,1024000000") #define eps 1e-10#define inf 0x3f3f3f3f#define PI pair<int, int> typedef long long LL;const int maxn = 100 + 5;int d[maxn], vis[maxn], w[maxn][maxn];int n;int dijkstra(int s) {memset(vis, 0, sizeof(vis));for(int i = 1; i <= n; ++i) d[i] = inf;d[s] = 0;for(int i = 0; i < n; ++i) {int x, m = inf;for(int y = 1; y <= n; ++y) if(!vis[y] && d[y] < m) m = d[x=y];vis[x] = 1;for(int y = 1; y <= n; ++y) if(w[x][y] < inf) d[y] = min(d[y], d[x] + w[x][y]);}int ans = 0;for(int i = 1; i <= n; ++i) {ans = max(ans, d[i]);}return ans;}int main() {while(scanf("%d", &n) == 1) {for(int i = 1; i <= n; ++i) for(int j = 1; j < n; ++j) {w[i][j] = inf;}char s[50];for(int i = 1; i <= n; ++i) {for(int j = 1; j < i; ++j) {scanf("%s", s);if(s[0] != 'x') {int dis;sscanf(s, "%d", &dis);w[i][j] = w[j][i] = dis;}}}printf("%d\n", dijkstra(1));}return 0;}

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