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【洛谷】P2910 [USACO08OPEN] Clear And Present Danger S 的题解

洛谷传送门

题解

明天就是 CSP 初赛了，蒟蒻在这里祝各位大佬 RP += INF

一个很裸的 Floyd。

• Floyd 板子

for(int k = 1; k <= n; k ++) {for(int i = 1; i <= n; i ++) {for(int j = 1; j <= n; j ++) {dis[i][j] = min(dis[i][k] + dis[k][j], dis[i][j]);}}}

用 Floyd 先求出两数的最短距离，然后跑一遍 $1$ 到 $n$ 最短路，输出答案即可。时间复杂度 $O(n^3)$。

代码

#include <bits/stdc++.h>
#define lowbit(x) x & (-x)
#define endl "\n"
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
namespace fastIO {inline int read() {register int x = 0, f = 1;register char c = getchar();while (c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();return x * f;}inline void write(int x) {if(x < 0) putchar('-'), x = -x;if(x > 9) write(x / 10);putchar(x % 10 + '0');return;}
}
using namespace fastIO;
int n, m, ans = 0, dis[105][105], a[10005];
int main() {//freopen(".in","r",stdin);//freopen(".out","w",stdout);n = read(), m = read();for(int i = 1; i <= m; i ++) {a[i] = read();}for(int i = 1; i <= n; i ++) {for(int j = 1; j <= n; j ++) {dis[i][j] = read();}}for(int k = 1; k <= n; k ++) {for(int i = 1; i <= n; i ++) {for(int j = 1; j <= n; j ++) {dis[i][j] = min(dis[i][k] + dis[k][j], dis[i][j]);}}}for(int i = 2; i <= m; i ++) {ans += dis[a[i - 1]][a[i]];}write(ans);return 0;
}