zoj 1298 && poj 1135 Domino Effect

来源：互联网发布：海牛大数据编辑：程序博客网时间：2024/06/02 07:08

题意：本题的要求是推倒第一张关键牌后，最后倒下的牌的位置及时间。

思路：

最后倒下的牌只有两种情形：⑴最后倒下的牌就是关键牌 ⑵最后倒下的牌是两张关键牌之间的普通牌。所以我们需要判断最后到底是那种情形。

1.计算每一张关键牌倒下的时间。取最大值maxtime1。

2.计算每一行完全倒下的时间。每一行倒下的时间t = (time[i] + time[j] + edge[i][j])/2.0,去最大值maxtime2。

3.如果maxtime2 > maxtime1 则最后倒下的牌是普通牌，否则是关键牌。

代码：

#include <iostream>#include <cstdio>using namespace std;const int inf = 10000000;const int maxn = 505;int edge[maxn][maxn];int Time[maxn];bool vis[maxn];int n,m;bool Input(){int u,v,w;cin>>n>>m;if(n == 0 && m == 0)return false;for(int i = 0; i <= n; i++){for(int j = 0; j <= n; j++)edge[i][j] = inf;}for(int i = 0; i < m; i++){cin>>u>>v>>w;edge[u][v] = edge[v][u] = w;}return true;}void Dijkstra(){for(int i = 1; i <= n; i++){vis[i] = 0; Time[i] = edge[1][i];}vis[1] = 1; Time[1] = 0;for(int i = 0; i < n-1; i++){int min = inf;int v;for(int j = 1; j <= n; j++){if(!vis[j] && Time[j] < min){min = Time[j]; v = j;}}vis[v] = 1;for(int j = 1; j <= n; j++){if(!vis[j] && Time[v] + edge[v][j] < Time[j]){Time[j] = Time[v] + edge[v][j];}}}}void Solve(){static int cnt = 0;Dijkstra();double maxtime1 = -inf; int pos;for(int i = 1; i <= n; i++){if(Time[i] > maxtime1){maxtime1 = Time[i]; pos = i;}}double maxtime2 = -inf; int pos1,pos2;for(int i = 1; i <= n; i++){for(int j = i+1; j <=n; j++){if(edge[i][j] == inf)continue;double t = (Time[i] + Time[j] + edge[i][j]) / 2.0;if(t > maxtime2){maxtime2 = t;pos1 = i;pos2 = j;}}}printf("System #%d\n",++cnt);printf("The last domino falls after ");if(maxtime2 > maxtime1){printf("%.1lf seconds, between key dominoes %d and %d.\n\n",maxtime2,pos1,pos2);}else{printf("%.1lf seconds, at key domino %d.\n\n",maxtime1,pos);}}int main(){while(Input()){Solve();}return 0;}

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