HDU 1599 find the mincost route 、 poj 1734 Sightseeing trip
阿新 • • 發佈:2019-02-09
今天無向圖求最小環的演算法,重溫了一遍floyd。
首先對於floyd演算法,是一種動態規劃的思想
dp[k][i][j]代表的含義是從ui到uj經過的除端點以外最大的不超過k的點的最小值
dp[k][i][j] = min(dp[k - 1][i][k] + dp[k - 1][k][j],dp[k][i][j]);
而求floyd時可以順便求得最小環,對於構成最小環的點集{x1, x2, ...,xk},其中必定存
在一個最大點xk,那麼最小環的長度可以由 dp[i][j]+dis[j][xk]+dis[xk][i]
來表
示,其中i,j均在點集中且小於xk,dp[i][j]是i到j的最短路徑(且最大經過的點為xk
-1)。對於每一個xk,列舉所有的dp[i][j](0 <= i , j <= xk-1)。便可求的最小環。
HDU 1599 find the mincost route
#include<iostream> #include<cstring> #include<cstdio> using namespace std; #define MAXN 105 #define INF 0x1f1f1f1f int dp[MAXN][MAXN],w[MAXN][MAXN],n,m; int floyd() { int ans = INF; for(int k = 1;k <= n;k++) { for(int i = 1;i < k;i++) for(int j = 1 + i;j < k;j++) ans = min(ans,dp[i][j] + w[i][k] + w[k][j]); for(int i = 1;i <= n;i++) for(int j = 1;j <= n;j++) dp[i][j] = min(dp[i][k] + dp[k][j],dp[i][j]); } return ans; } int main() { int u,v,val; while(scanf("%d%d",&n,&m) != EOF) { memset(dp,0x1f,sizeof(dp)); memset(w,0x1f,sizeof(w)); while(m--) { scanf("%d%d%d",&u,&v,&val); dp[u][u] = dp[v][v] = w[u][u] = w[v][v] = 0; w[v][u] = w[u][v] = dp[u][v] = dp[v][u] = min(val,dp[u][v]); } int ans = floyd(); if(ans == INF)puts("It's impossible."); else printf("%d\n",ans); } }
poj 1734 Sightseeing trip
#include<iostream> #include<cstring> #include<cstdio> using namespace std; #define MAXN 105 #define INF 0x1f1f1f1f int dp[MAXN][MAXN],w[MAXN][MAXN],pre[MAXN][MAXN],arr[MAXN]; int n,m,len; void dfs(int i,int j) { int k = pre[i][j]; if(k == -1) { arr[len++] = i; return; } dfs(i,k),dfs(k,j); } void floyd() { int ans = INF; len = 0; for(int k = 1;k <= n;k++) { for(int i = 1;i < k;i++) for(int j = 1 + i;j < k;j++) { int cur = dp[i][j] + w[i][k] + w[k][j]; if(cur < ans) { ans = cur; len = 0; arr[len++] = k; dfs(i,j); arr[len++] = j; } } for(int i = 1;i <= n;i++) for(int j = 1;j <= n;j++) { if(dp[i][j] > dp[i][k] + dp[k][j]) { dp[i][j] = dp[i][k] + dp[k][j]; pre[i][j] = k; } } } } int main() { int u,v,val; while(scanf("%d%d",&n,&m) != EOF) { memset(dp,0x1f,sizeof(dp)); memset(w,0x1f,sizeof(w)); memset(pre,-1,sizeof(pre)); while(m--) { scanf("%d%d%d",&u,&v,&val); dp[u][u] = dp[v][v] = w[u][u] = w[v][v] = 0; w[v][u] = w[u][v] = dp[u][v] = dp[v][u] = min(val,dp[u][v]); } floyd(); if(len < 2)puts("No solution."); else for(int i = 0;i < len;i++) if(i != len - 1)printf("%d ",arr[i]); else printf("%d\n",arr[i]); } }