雕刻时光

摘要： 区间覆盖最小代价由于N<=10000排序后想到N*N的遍历不断更新add[i](表示到i最小代价)里的值，使之最小View Code #include<stdio.h>#include<iostream>#include<algorithm>using namespace std;const int maxint=999999999;struct data{ int ll,rr,c;}s[10009];int add[10009];int cmp(data a,data b){ if(a.ll==b.ll) return a.rr<b.rr; e 阅读全文

摘要： DIJK水过View Code #include<stdio.h>#include<string.h>#include<stack>#include<iostream>using namespace std;int n,m;int map[1009][1009];int path[1009];int dis[1009];bool used[1009];const int maxint=999999999;void dijk(){ memset(path,-1,sizeof(path)); memset(used,0,sizeof(used)); 阅读全文