标签:ini ++ contest max 取出 direction blank get one
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0 2 1 1 -1HintIf you need a larger stack size, please use #pragma comment(linker, "/STACK:102400000,102400000") and submit your solution using C++.
求最短路:把一个集合的点看做是一个点,这样就能够用djstra算法做了。然后因为每一个点最多标记一次最短路,用set维护一个点集合。
当最短路找到一个一个集合的时候,把这个集合里还存在的点都取出就可以。取出后。每一个点又能够去两个集合。
再向保存最短路的set里更新集合信息就可以。具体看代码。
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<set> using namespace std; #define maxn 200007 #define ll long long int lp[maxn],rp[maxn]; ll cosw[maxn]; ll dist[maxn]; set<int> haha; struct Node{ int id; ll cost; }; bool operator < (Node a,Node b){ if(a.cost == b.cost) return a.id < b.id; return a.cost < b.cost; } set<Node> mind; int main(){ int t,n; scanf("%d",&t); while(t--){ scanf("%d",&n); for(int i = 0;i < n; i++) scanf("%d",&lp[i]); for(int i = 0;i < n; i++) scanf("%d",&rp[i]); for(int i = 0;i < n; i++) scanf("%d",&cosw[i]); haha.clear(); mind.clear(); memset(dist,-1,sizeof(dist)); dist[0] = 0; Node x,y; x.id = 0; x.cost = cosw[0]; mind.insert(x); for(int i = 1;i < n; i++) haha.insert(i); set<int>::iterator it,it2; while(mind.size() > 0){ x = *mind.begin(); mind.erase(mind.begin()); it = haha.lower_bound(x.id - rp[x.id]); while(it != haha.end() && *it <= x.id - lp[x.id]){ y.id = *it; y.cost = x.cost + cosw[y.id]; dist[y.id] = x.cost; mind.insert(y); it2 = it++; haha.erase(it2); } it = haha.lower_bound(x.id + lp[x.id]); while(it != haha.end() && *it <= x.id + rp[x.id]){ y.id = *it; y.cost = x.cost + cosw[y.id]; dist[y.id] = x.cost; mind.insert(y); it2 = it++; haha.erase(it2); } } for(int i = 0;i < n; i++){ if(i) printf(" "); printf("%I64d",dist[i]); } printf("\n"); } return 0; }
标签:ini ++ contest max 取出 direction blank get one
原文地址:http://www.cnblogs.com/wzzkaifa/p/6918939.html