Floyd 求最短路(poj 1161)
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题目:Walls
题意:给定一个图,求其中几个点相连最少要穿越的边数。
思路:这题的图要重新建,不能用原图,新图是这样的:将一个圈化为点,之间的关系是两个圈是否有公共边,然后就是求最短路问题了;
#include <iostream> #include <algorithm> #include <stdlib.h> #include <time.h> #include <cmath> #include <cstdio> #include <string> #include <cstring> #include <vector> #include <queue> #include <stack> #include <set> #define c_false ios_base::sync_with_stdio(false); cin.tie(0) #define INF 0x3f3f3f3f #define INFL 0x3f3f3f3f3f3f3f3f #define zero_(x,y) memset(x , y , sizeof(x)) #define zero(x) memset(x , 0 , sizeof(x)) #define MAX(x) memset(x , 0x3f ,sizeof(x)) #define swa(x,y) {LL s;s=x;x=y;y=s;} using namespace std ; #define N 500 const double PI = acos(-1.0); typedef long long LL ; int mapp[N][N], region[N][N], p_num[N], n, m, r; struct Person{ int adr; int t; ///所在环数; int reg[N]; ///环 }person[N]; int find_dis(int i, int j){ ///寻找i,j 两区域间的边,若没有,返回-1 int ii, jj; for(ii = 1; ii <= p_num[i]; ii++){ for(jj = 1; jj <= p_num[j]; jj++){ if(region[i][ii] == region[j][jj]){ if(region[i][ii+1] == region[j][jj+1] || region[i][ii+1] == region[j][jj-1] || region[i][ii-1] == region[j][jj+1] || region[i][ii-1] == region[j][jj-1]) return 1; } } } return -1; } void Floyd(){ ///寻找最短路 int i, j, k; for(k = 1; k<= m; k++) for(i = 1; i<= m; i++) for(j = 1; j<= m; j++) mapp[i][j] = mapp[j][i] = min(mapp[i][j], mapp[i][k]+mapp[k][j]); } int search_(int i){ //查找i区域到俱乐部每个成员间的距离和; int j, k, s = 0; for(j = 1; j<= r;j++){ int d = INF; for(k = 1; k <=person[j].t; k++) d = min(d, mapp[person[j].reg[k]][i]); s+=d; } return s; } int main(){ //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int i,j,k; while(~scanf("%d",&m)){ scanf("%d%d",&n,&r); for(i = 1; i<= r;i++){ scanf("%d",&person[i].adr); person[i].t = 0; } for(i = 1; i <= m; i++){ scanf("%d", &p_num[i]); for(j = 1; j <= p_num[i]; j++){ scanf("%d", ®ion[i][j]); for(k = 1; k <= r; k++){ if(person[k].adr == region[i][j]) person[k].reg[ ++person[k].t] = i; } } region[i][0] = region[i][ p_num[i] ]; region[i][p_num[i]+1] = region[i][1]; } for(i = 1; i <= m; i++){ for(j = 1; j<=m; j++){ if(i == j){ mapp[i][j] = mapp[j][i] = 0;continue; } int ans = find_dis(i, j); if(ans == 1) mapp[i][j] = mapp[j][i] =1; else mapp[i][j] = mapp[j][i] =INF; } } Floyd(); int min_dis = search_(1); for(i = 2; i <=m ;i++){ int d =search_(i); min_dis = min(d, min_dis); } printf("%d\n",min_dis); } return 0; }
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