HDU 3665 Seaside （最短路，Floyd）

题意：给定一个图，你家在0，让你找出到沿海的最短路径。

析：由于这个题最多才10个点，那么就可以用Floyd算法，然后再搜一下哪一个是最短的。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <stack>
using namespace std;
 
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 10 + 5;
const int mod = 1e9;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
inline int Min(int a, int b){  return a < b ? a : b; }
inline int Max(int a, int b){  return a > b ? a : b; }
int d[maxn][maxn];
int a[maxn];
 
int solve(){
    for(int i = 0; i < n; ++i)
        for(int j = 0; j < n; ++j)
            for(int k = 0; k < n; ++k)
                d[i][j] = Min(d[i][j], d[i][k]+d[k][j]);
    int ans = INF;
    for(int i = 0; i < n; ++i)
        if(a[i] == 1)  ans = Min(ans, d[0][i]);
    return ans;
}
 
int main(){
    while(scanf("%d", &n) == 1){
        memset(a, -1, sizeof(a));
        int x, y;
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < n; ++j)  d[i][j] = INF;
 
        for(int i = 0; i < n; ++i){
            scanf("%d %d", &m, &a[i]);
            for(int j = 0; j < m; ++j){
                int u, w;
                scanf("%d %d", &u, &w);
                d[i][u] = w;
            }
        }
 
        int ans = solve();
        printf("%d\n", ans);
    }
    return 0;
}