题意:给出一个又向图每个图有权值和编号(正方形里的是编号),从第0号节点开始每次向当前节点所连的点中权值最大的节点移动(不会存在权值相同的节点),问最后所在的节点编号和经过的节点的权值之和。
解:模拟就好~
1 #include<cstdio> 2 #include<algorithm> 3 #include<cmath> 4 #include<cstring> 5 #include<vector> 6 7 using namespace std; 8 9 int T; 10 int n,m; 11 int f[107]; 12 vector <int> G[107]; 13 14 int main(){ 15 scanf("%d",&T); 16 for (int cas=1;cas<=T;cas++){ 17 scanf("%d%d",&n,&m); 18 for (int i=0;i<n;i++){ 19 scanf("%d",&f[i]); 20 G[i].clear(); 21 } 22 for (int i=0;i<m;i++){ 23 int x,y; 24 scanf("%d%d",&x,&y); 25 G[x].push_back(y); 26 } 27 int now=0; 28 int totv=0; 29 while (1){ 30 // printf("%d %d\n",now,totv); 31 int to=0; 32 int tov=0; 33 for (int i=0;i<G[now].size();i++){ 34 if (f[G[now][i]]>tov){ 35 to=G[now][i]; 36 tov=f[G[now][i]]; 37 } 38 } 39 if (to==0) break; 40 totv+=tov; 41 now=to; 42 } 43 printf("Case %d: %d %d\n",cas,totv,now); 44 } 45 return 0; 46 } 47 /* 48 2 49 6 6 50 0 8 9 2 7 5 51 5 4 52 5 3 53 1 5 54 0 1 55 0 2 56 2 1 57 6 6 58 0 8 9 2 6 5 59 5 4 60 5 3 61 1 5 62 0 1 63 0 2 64 2 1 65 */
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原文:http://www.cnblogs.com/baby-mouse/p/4296292.html