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[N][N];
int dis[N], vis[N];

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[N][N];
int vis[N], dis[N];
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;
}