2012 Maryland High School Battleground Preservation map容器+Floyd
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题目链接地址 第五题
当时不知怎么~~竟然让羽哥写了dijkstra~求两源的最短路~~其实还可以用Floyd来写 多源的最短路~~
下面是自己写的Flody +map代码:
#include<stdio.h>#include<map> #include<iostream>using namespace std;#define INF 1000000000int pic[111][111];void init(int n){ //队邻接矩阵进行初始化 for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) if(i==j)pic[i][j]=0; else pic[i][j]=INF; }void Floyd(int n){ //用FLOYD求解 for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) for(int k=1;k<=n;k++) pic[j][k]=min(pic[j][i]+pic[i][k],pic[j][k]); }int main(){ //freopen("1.txt","r",stdin); //freopen("2.txt","w",stdout); int x; scanf("%d",&x); while (x--){ map <string,int> mapp; int n,m,len,ji=1; string a,b; scanf("%d%d",&n,&m); init(n); for (int i=1;i<=m;i++){ cin >>a >>b >>len; //cout<<a<<b<<len<<endl; if (mapp.find(a)==mapp.end()){ mapp[a]=ji++; } if (mapp.find(b)==mapp.end()){ mapp[b]=ji++; } pic[mapp[a]][mapp[b]]=len; } Floyd(n); cin >>a >>b; if( ( mapp.find(a)==mapp.end() ) || ( mapp.find(a)==mapp.end() ))printf("FIGHT!\n"); //若有一个没有通,就说明不存在两点的连线。直接输出fight else { int ansa=pic[mapp[a]][mapp[b]],ansb=pic[mapp[b]][mapp[a]]; if(ansa>ansb)cout<<b<<endl; else if(ansa<ansb) cout<<a<<endl; else cout<<"FIGHT!"<<endl; } } return 0;}
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