（kuangbin带你飞--最短路径）Heavy Transportation

原题目BackgroundHugo 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.ProblemYou 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 (Hugos place) to crossing n (the customers place). You may assume that there is at least one path. All streets can be travelled in both directions.InputThe 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.OutputThe 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 Input1 3 3 1 2 3 1 3 4 2 3 5Sample OutputScenario #1: 4从点1到点n很多条路径每条路径都有一个最小的权值求这些最小的权值里的最大值。就是不断维护d数组#includestdio.h #includealgorithm #includeiostream #includestring.h using namespace std; const int M1005; int t,n,m,cas0; int mmp[M][M],vis[M],d[M]; //vis数组用来判断是否查询过 //mmp数组储存图 //d数组表示1到N的最大距离 void dijk() { memset(vis,0,sizeof(vis)); for(int i1;in;i) d[i]mmp[1][i]; d[1]0; vis[0]1; for(int i1;in;i) { int u-1,m0;//m表示此时i到各个点的最大距离 for(int j1;jn;j) { if(!vis[j]d[j]m) { md[j]; uj; } } vis[u]1; for(int i1;in;i) { if(mmp[u][i]) d[i]max(d[i],min(d[u],mmp[u][i])); } } } int main() { scanf(%d,t); while(t--) { memset(mmp,0,sizeof(mmp)); int u,v,w; scanf(%d%d,n,m); for(int i1;im;i) { scanf(%d%d%d,u,v,w); mmp[u][v]mmp[v][u]w; } dijk(); if(cas) puts(); printf(Scenario #%d:\n,cas); printf(%d\n,d[n]); } }这道题目要求的是从1到N各个路径最小值的最大值即d[y]max(d[y],min(d[x],m[x][y])).