Poj 1797 Heavy Transportation ( 最短路变形

Heavy Transportation

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 the 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 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 travelled in both directions.

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 containing “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

这里写代码片Scenario #1:4

题意

要从城市1到城市N运送货物，有M条道路，每条道路都有它的最大载重量，问从城市1到城市N运送最多的重量是多少。

题解：

可以看成最短路模型 不过松弛条件需要改下
我们就需要去选取离源点权值最大的点,使得它的该路径的最大载重量大一些;

AC代码

#include <cstdio>#include <cmath>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define LL long long#define CLR(a,b) memset(a,(b),sizeof(a))const int INF = 0x3f3f3f3f;const LL INFLL = 0x3f3f3f3f3f3f3f3f;const int N = 1e3+10;int  mps[N][N];       //邻接矩阵，表示从->的距离int dis[N];       //dst[i] 从到i的距离bool vis[N];      // 标记节点是否被访问int n, m;void init(){    for(int i = 1; i <= n; i++) {        for(int j = 1;j <= n; j++) mps[i][j] = -1;    }}void Dijkstra(int s){    CLR(vis,false);    for(int i = 1;i <= n; i++) {        dis[i] =  mps[s][i];    }    vis[s] = true;    int minx;    int k;    for(int i = 1; i <= n; i++){        minx = -1;        for(int j = 1; j <= n; j++) {            if(!vis[j] && dis[j]>minx) {                minx = dis[j];                k = j;            }        }        vis[k] = true;        for(int j = 1;j <= n; j++) {            if(!vis[j] && mps[j][k]!=-1) {                if(dis[j] == -1) dis[j] = min(mps[k][j],dis[k]);                else dis[j] = max(dis[j],min(dis[k],mps[k][j]));            }        }    }}int main(){    int zz = 0;    int T;    scanf("%d",&T);    while(T--) {        scanf("%d%d",&n,&m);        init();        for(int i = 1;i <= m; i++) {            int x, y, z;            scanf("%d%d%d",&x,&y,&z);            mps[x][y] = mps[y][x] = z;        }        Dijkstra(1);        printf("Scenario #%d:\n",++zz);        printf("%d\n",dis[n]);        printf("\n");    }return 0;}