标签:技术 block com tor hid images closed pac amp
题目链接:51nod 1445 变色DNA
看了相关讨论再去用最短路:val[i][j]之间如果是‘Y’,说明i可以到达j,并且i到达j的代价是i那行 1到j-1 里面‘Y’的数量。
最后,求 0到n-1的最短路。
感觉读懂了题意就真的简单了。。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<vector> 5 #include<queue> 6 #define CLR(a,b) memset((a),(b),sizeof((a))) 7 using namespace std; 8 const int inf = 0x3f3f3f3f; 9 const int N = 51; 10 int d[N], vis[N]; 11 char mp[N]; 12 int n; 13 struct qnode{ 14 int v, c; 15 qnode(int _v = 0,int _c = 0):v(_v),c(_c) {} 16 bool operator < (const qnode &r)const{ 17 return r.c < c; 18 } 19 }; 20 vector<int>E[N]; 21 void dij(){ 22 priority_queue<qnode>q; 23 for(int i = 1; i <= n; ++i){ 24 d[i] = inf; 25 vis[i] = 0; 26 } 27 d[1] = 0; 28 q.push(qnode(1, 0)); 29 while(!q.empty()){ 30 qnode t = q.top(); q.pop(); 31 int u = t.v; 32 if(vis[u]) 33 continue; 34 vis[u] = 1; 35 for(int i = 0; i < E[u].size(); ++i){ 36 int v = E[u][i]; 37 if(!vis[v] && d[u] + i < d[v]){ 38 d[v] = d[u] + i; 39 q.push(qnode(v, d[v])); 40 } 41 } 42 } 43 } 44 int main(){ 45 int t, i, j; 46 scanf("%d", &t); 47 while(t--){ 48 scanf("%d", &n); 49 for(i = 0; i < N; ++i) E[i].clear(); 50 for(i = 1; i <= n; ++i){ 51 scanf("%s",mp+1); 52 for(j = 1; j <= n; ++j) 53 if(mp[j] == ‘Y‘) 54 E[i].push_back(j); 55 } 56 dij(); 57 printf("%d\n", d[n]==inf?-1:d[n]); 58 } 59 return 0; 60 }
标签:技术 block com tor hid images closed pac amp
原文地址:http://www.cnblogs.com/GraceSkyer/p/5988116.html