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隐式的图搜索,存不下边,所以只有枚举转移就行了,转移的时候判断合法可以用位运算优化,
二进制pre[i][0]表示可以出现的bug,那么u&pre[i][0] == u就表示u是可以出现的bug集合的子集,
pre[i][1]表示必须出现的bug,那么u|pre[i][i] != u表示把必须出现的bug添加到u中,u中bug增加,这是不合法的。
正权最短路就dijkstra,用spfa以前某题狂T有阴影。被输出格式坑得不要不要的,如果是if(kas) putchar(‘\n‘);就会WA...
#include<bits/stdc++.h> using namespace std; const int maxm = 100; const int maxn = 20; int pre[maxm][2],nxt[maxm][2]; int cost[maxm]; int n,m; int dist[1<<maxn]; typedef pair<int,int> Node; #define fi first #define se second //bitset<20> temp; #define bug(u)\ temp = u; cout<<#u<<‘=‘<<temp<<endl; #define cer(x)\ cout<<"dist="<<x<<endl; const int INF = 0x3f3f3f3f; void dijkstra() { priority_queue<Node,vector<Node>,greater<Node> > q; memset(dist,0x3f,sizeof(int)*(1<<n)); q.push(Node(0,(1<<n)-1)); dist[(1<<n)-1] = 0; while(q.size()){ Node x = q.top(); q.pop(); if(x.se == 0) { printf("Fastest sequence takes %d seconds.\n",dist[0]); return; } if(x.fi != dist[x.se]) continue; int u = x.se; for(int i = 0; i < m; i++){ if( (pre[i][0]&u) == u && (pre[i][1]|u) == u){ int v = (u&nxt[i][0])|nxt[i][1]; if(dist[v] > dist[u]+cost[i]){ dist[v] = dist[u] + cost[i]; q.push(Node(dist[v],v)); } } } } puts("Bugs cannot be fixed."); } int main() { //freopen("in.txt","r",stdin); int kas = 0; char s1[maxn+5],s2[maxn+5]; while(scanf("%d%d",&n,&m),n){ for(int i = 0; i < m ; i++){ scanf("%d%s%s",cost+i,s1,s2); nxt[i][0] = nxt[i][1] = pre[i][0] = pre[i][1] = 0; for(int j = 0; j < n; j++){ if(s1[j] == ‘+‘) pre[i][1] |= 1<<j; if(s1[j] != ‘-‘) pre[i][0] |= 1<<j; if(s2[j] == ‘+‘) nxt[i][1] |= 1<<j; if(s2[j] != ‘-‘) nxt[i][0] |= 1<<j; } } printf("Product %d\n",++kas); dijkstra(); putchar(‘\n‘); } return 0; }
UVA - 658 It's not a Bug, it's a Feature!
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原文地址:http://www.cnblogs.com/jerryRey/p/4758954.html