最短路变形 poj3615&

问题：

牛要跨过一些障碍，希望以最小的体力跨过障碍，并且对于一条路径，只在乎其中最高的障碍。

输入N代表站点数，标记为1—N，输入M代表路径数，从站点S到E之间需要跨过高度为H的障碍。

输入T代表牛要完成的任务数。对于每个任务，输入A，B，输出一条从站点A到B的路径，使需要跨过的最高障碍为最低。

代码：（Floyd

/*******************************************
Problem: 3615		User:
Memory: 960K		Time: 688MS
Language: G++		Result: Accepted
********************************************/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 305;

int mp

;
int dis
, vis
;

void floyd(int n)
{
int i, j, k;
for (k = 1; k <= n; ++k)
for (i = 1; i <= n; ++i)
for (j = 1; j <= n; ++j)
if (mp[i][j] > max(mp[i][k] , mp[k][j]))
mp[i][j] = max(mp[i][k], mp[k][j]);
}

int main()
{
int n, m, t;
while (scanf("%d%d%d", &n, &m, &t) != EOF) {
int i, j;
int a, b, h;
for (i = 1; i <= n; ++i)
for (j = 1; j <= n; ++j)
mp[i][j] = INF;
for (i = 0; i < m; ++i) {
scanf("%d%d%d", &a, &b, &h);
mp[a][b] = h;
}
floyd(n);
for (i = 0; i < t; ++i) {
scanf("%d%d", &a, &b);
printf("%d\n", mp[a][b] == INF ? -1 : mp[a][b]);
}
}
return 0;
}

　　

poj2263

和上题相似，求两点之间最小值最大的路径

代码：（dijkstra

/******************************************
Problem: 2263		User:
Memory: 812K		Time: 125MS
Language: G++		Result: Accepted
*******************************************/
#include <iostream>
#include <cstring>
#include <string>
#include <map>
#include <algorithm>
#include <cstdio>
using namespace std;

const int N = 205;
const int INF = 0x3f3f3f3f;

int mp

;
int vis
, dis
;
map<string, int>my_map;

int dijkstra(int n, int s, int e)
{
int i, j;
memset(vis, 0, sizeof(vis));
for (i = 1; i <= n; ++i)
dis[i] = mp[s][i];
vis[s] = 1;
for (i = 0; i < n; ++i) {
int max_dis = 0;
int max_x = 1;
for (j = 1; j <= n; ++j) {
if (!vis[j] && dis[j] > max_dis) {
max_dis = dis[j];
max_x = j;
}
}
if (max_x == e) return max_dis;
vis[max_x] = 1;
for (j = 1; j <= n; ++j) {
if (!vis[j] && dis[j] < min(dis[max_x], mp[j][max_x]))
dis[j] = min(dis[max_x], mp[j][max_x]);
}
}
return 0;
}

int main()
{
int n, m;
int times = 1;
while (scanf("%d%d", &n, &m) != EOF) {
if (n == 0 && m == 0) break;
int i, j;
my_map.clear();
int cnt = 1;
string stra, strb;
int x, y;
int weight;
for (i = 1; i <= n; ++i)
for (j = 1; j <= n; ++j)
mp[i][j] = 0;

for (i = 0; i < m; ++i) {
cin >> stra >> strb >> weight;
if (my_map[stra] == 0)  my_map[stra] = cnt++;
if (my_map[strb] == 0)  my_map[strb] = cnt++;
x = my_map[stra];
y = my_map[strb];
if (mp[x][y] < weight)
mp[y][x] = mp[x][y] = weight;
}
cin >> stra >> strb;
x = my_map[stra];
y = my_map[strb];
printf("Scenario #%d\n%d tons\n\n", times++, dijkstra(n, x, y));

}
return 0;
}