HDU 5137 How Many Maos Does the Guanxi Worth（最短路+枚舉）

題解：

首先先跑一遍最短路，用pre數組記錄最短路路徑，枚舉路徑上的所有點（不包括 1 和 n）,標記這些點再來跑最短路記錄其中最大的最短路。

#include <algorithm>
#include  <iostream>
#include   <cstdlib>
#include   <cstring>
#include    <cstdio>
#include    <string>
#include    <vector>
#include    <bitset>
#include     <stack>
#include     <cmath>
#include     <deque>
#include     <queue>
#include      <list>
#include       <set>
#include       <map>
#define mem(a, b) memset(a, b, sizeof(a))
#define pi acos(-1)
using namespace std;
typedef long long ll;
const int  inf = 0x3f3f3f3f;
const int maxn = 35;

struct node{
	int v, c, use;
	node(){};
	node(int v, int c, int use):v(v),c(c), use(use){};
};

struct node1{
	int v, c;
	node1(){};
	node1(int v, int c):v(v),c(c){};
}pre[maxn];

vector<node> v[maxn];
int vis[maxn], lowcost[maxn], mark[maxn];

void Dijkstra(int s, int n, int op){
	for(int i = 0; i <= n; i++){
		lowcost[i] = inf;
		vis[i] = 0;
		if(op == 0){
			pre[i] = node1(-1, inf);
		}
	}
	lowcost[s] = 0;
	for(int i = 0; i < n; i++){
		int k = -1;
		int minn = inf;
		for(int j = 1; j <= n; j++){
			if(!vis[j] && lowcost[j] < minn){
				minn = lowcost[j];
				k = j;
			}
		}
		if(k == -1){
			break;
		}
		vis[k] = 1;
		for(int j = 0; j < v[k].size(); j++){
			if(mark[v[k][j].v] && !vis[v[k][j].v] && lowcost[k] + v[k][j].c < lowcost[v[k][j].v]){
				lowcost[v[k][j].v] = lowcost[k] + v[k][j].c;
				if(op == 0){
					pre[v[k][j].v] = node1(k, v[k][j].c);
				}
			}
		}
	}
}

int main(){
	int n, m;
	while(scanf("%d %d", &n, &m), n&&m){
		for(int i = 1; i <= n; i++){
			v[i].clear();
			mark[i] = 1;
		}
		for(int i = 0; i < m; i++){
			int x, y, c;
			scanf("%d %d %d", &x, &y, &c);
			v[x].push_back(node(y, c, 0));
			v[y].push_back(node(x, c, 0));
		}
		Dijkstra(1, n, 0);
		int ans = lowcost[n];
		int s = n;
		while(pre[s].v != 1){
			mark[pre[s].v] = 0;
			Dijkstra(1, n, 1);
			ans = max(ans, lowcost[n]);
			mark[pre[s].v] = 1;
			s = pre[s].v;
		}
		if(ans == inf){
			printf("Inf\n");
		}
		else{
			printf("%d\n", ans);
		}
	} 
}