poj -- 2288 Islands and Bridges
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题意:给你n个岛屿,m条路,每个岛屿都有一个值vi,经过这个岛屿会得到这个值,规则:
1、经过所有的岛屿 sum += vi;
2、连续两个岛屿(即连通) sum += vi * vj;
3、连续三个岛屿(即两两连通)sum += vi + vj + vk;
求得到最大的值,并且问的到最大的值有多少种路线;(最后的路线/2,无向图)
dp[i][j][k] 表示第i个状态 j-->k 得到的最大值
s[i][j][k] 表示第i个状态 j-->k 得到的最大值的路线的数量
注:1001 ---> 1101 1的位置与0的位置和3的位置有关
#include<cstdio>#include<cmath>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef __int64 LL;#define M 1<<14LL dp[M][14][14];LL s[M][14][14];LL v[14];int map[14][14];int n,m;void DP(){ int cnt = 1 << n; for(int i = 0;i < cnt;i++) { for(int j = 0;j < n;j++) { for(int k = 0;k < n;k++) { if(dp[i][j][k] == -1) continue; for(int r = 0;r < n;r++) { if(r == j || r == k || i&(1<<r) || !map[k][r]) continue; LL xx = v[r] + v[r] * v[k] + dp[i][j][k]; if(map[j][r]) xx += v[j] * v[k] * v[r]; LL& Vmax = dp[i+(1<<r)][k][r]; if(Vmax < xx){ Vmax = xx; s[i+(1<<r)][k][r] = s[i][j][k]; } else if(Vmax == xx) s[i+(1<<r)][k][r] += s[i][j][k]; } } } } LL ans = 0; LL num = 0; for(int i = 0;i < n;i++) { for(int j = 0;j < n;j++) { if(dp[cnt-1][i][j] == -1) continue; if(dp[cnt-1][i][j] > ans) { ans = dp[cnt-1][i][j]; num = s[cnt-1][i][j]; } else if(dp[cnt-1][i][j] == ans) num += s[cnt-1][i][j]; } } if(num) cout << ans << " " << num/2 << endl; else cout << "0 0" << endl;}int main(){ int t,u,w; cin >> t; while(t--) { cin >> n >> m; for(int i = 0;i < n;i++) cin >> v[i]; memset(map,0,sizeof(map)); memset(dp,-1,sizeof(dp)); memset(s,0,sizeof(s)); for(int i = 0;i < m;i++) { cin >> u >> w; u -- ; w --; map[u][w] = map[w][u] = 1; int st = (1 << u) + (1 << w); dp[st][u][w] = dp[st][w][u] = v[u] + v[w] + v[u] * v[w] ; s[st][u][w] = s[st][w][u] = 1; } if(n == 1){ cout << v[0] << " 1" << endl; continue; } DP(); }}
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