Poj1797 Heavy Transportation [Dijkstra]

Heavy Transportation

Time limit:3000 Ms		Memory limit:30000 K
Total submissions:21037		Accepted:5569

Description

Background
Hugo heavy is happy. after the breakdown of the cargolifter project he can now expand business. but he needs a clever man who tells him whether there really is a way from the place his customer has build his giant steel crane to the place where it is needed on which all streets can carry weight.
Fortunately he already has a plan of the city with all streets and bridges and all the allowed weights. unfortunately he has no idea how to find the maximum weight capacity in order to tell his customer how heavy the crane may become. but you surely know.

Problem
You are given the plan of the city, described by the streets (with weight limits) between the crossings, which are numbered from 1 to n. your task is to find the maximum weight that can be transported from crossing 1 (Hugo's place) to crossing N (the customer's place ). you may assume that there is at least one path. all streets can be traveled in both directions ctions.

Input

The first line contains the number of scenarios (city plans ). for each city the number n of street crossings (1 <= n <= 1000) and number M of streets are given on the first line. the following M lines contain triples of integers specifying start and end crossing of the street and the maximum allowed weight, which is positive and not larger than 1000000. there will be at most one street between each pair of crossings.

Output

The output for every scenario begins with a line ining "Scenario # I:", where I is the number of the scenario starting at 1. then print a single line containing the maximum allowed weight that Hugo can transport to the customer. terminate the output for the scenario with a blank line.

Sample Input

13 31 2 31 3 42 3 5

Sample output

Scenario #1:4

Source

Tud Programming Contest 2004, Darmstadt, Germany question: Calculate the minimum cut from 1 to n. First try DFS and the result is TLE.

#include <stdio.h>#include <string.h>#define maxn 1010#define maxm maxn * maxn#define inf 0x3f3f3f3fint head[maxn], n, m, id, ans, cas = 1;struct Node {    int v, c, next;} E[maxm];bool vis[maxn];void addEdge(int u, int v, int c) {    E[id].v = v; E[id].c = c;    E[id].next = head[u]; head[u] = id++;    E[id].v = u; E[id].c = c;    E[id].next = head[v]; head[v] = id++;}void getMap() {    int u, v, c; id = 0;    scanf("%d%d", &n, &m);    memset(head, -1, sizeof(int) * (n + 1));    while(m--) {        scanf("%d%d%d", &u, &v, &c);        addEdge(u, v, c);    }}void DFS(int k, int dis) {    if(k == n) {        if(dis > ans) ans = dis;        return;    }    for(int i = head[k]; i != -1; i = E[i].next) {                if(!vis[E[i].v]) {            int pre = dis;            vis[E[i].v] = 1;            if(E[i].c < dis) dis = E[i].c;            DFS(E[i].v, dis);            dis = pre; vis[E[i].v] = 0;        }    }}void solve() {    ans = 0;    memset(vis, 0, sizeof(bool) * (n + 1));    vis[1] = 1; DFS(1, inf);    printf("Scenario #%d:\n%d\n\n", cas++, ans);}int main() {    // freopen("stdin.txt", "r", stdin);    int t;    scanf("%d", &t);    while(t--) {        getMap();        solve();    }    return 0;}

Then I tried Dijkstra and ran the .. dis array to store the minimum cut from the current point to the source point.

#include <stdio.h>#include <string.h>#define maxn 1010#define maxm maxn * maxn#define inf 0x3f3f3f3fint head[maxn], n, m, id, ans, cas = 1;struct Node {    int v, c, next;} E[maxm];int dis[maxn];bool vis[maxn];int max(int a, int b) {    return a > b ? a : b;}int min(int a, int b) {    return a < b ? a : b;}void addEdge(int u, int v, int c) {    E[id].v = v; E[id].c = c;    E[id].next = head[u]; head[u] = id++;    E[id].v = u; E[id].c = c;    E[id].next = head[v]; head[v] = id++;}void getMap() {    int u, v, c; id = 0;    scanf("%d%d", &n, &m);    memset(head, -1, sizeof(int) * (n + 1));    while(m--) {        scanf("%d%d%d", &u, &v, &c);        addEdge(u, v, c);    }}int getNext() {    int pos = -1, val = 0;     for(int i = 1; i <= n; ++i)        if(dis[i] > val && !vis[i]) {            val = dis[i]; pos = i;        }    return pos;}void Dijkstra(int start, int end) {    memset(dis, 0, sizeof(int) * (n + 1));    dis[start] = inf;    int i, u = start, v;    while(u != -1) {        vis[u] = 1;        if(u == end) return;        for(i = head[u]; i != -1; i = E[i].next) {            if(!vis[v = E[i].v]) dis[v] = max(dis[v], min(E[i].c, dis[u]));        }        u = getNext();    }   }void solve() {    memset(vis, 0, sizeof(bool) * (n + 1));    Dijkstra(1, n);    printf("Scenario #%d:\n%d\n\n", cas++, dis[n]);}int main() {    // freopen("stdin.txt", "r", stdin);    int t;    scanf("%d", &t);    while(t--) {        getMap();        solve();    }    return 0;}

Poj1797 Heavy Transportation [Dijkstra]