uvalive5818 uva12376 As Long as I Learn, I Live
2015-04-21 18:24
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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~
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解:模拟就好~
#include<cstdio> #include<algorithm> #include<cmath> #include<cstring> #include<vector> using namespace std; int T; int n,m; int f[107]; vector <int> G[107]; int main(){ scanf("%d",&T); for (int cas=1;cas<=T;cas++){ scanf("%d%d",&n,&m); for (int i=0;i<n;i++){ scanf("%d",&f[i]); G[i].clear(); } for (int i=0;i<m;i++){ int x,y; scanf("%d%d",&x,&y); G[x].push_back(y); } int now=0; int totv=0; while (1){ // printf("%d %d\n",now,totv); int to=0; int tov=0; for (int i=0;i<G[now].size();i++){ if (f[G[now][i]]>tov){ to=G[now][i]; tov=f[G[now][i]]; } } if (to==0) break; totv+=tov; now=to; } printf("Case %d: %d %d\n",cas,totv,now); } return 0; } /* 2 6 6 0 8 9 2 7 5 5 4 5 3 1 5 0 1 0 2 2 1 6 6 0 8 9 2 6 5 5 4 5 3 1 5 0 1 0 2 2 1 */
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