President's Path CodeForces - 416E (最短路,计数)

大意: 给定无向图, 求任意两点间所有最短路经过的边数

刚开始想先用floyd跑出最短路, 然后在DAG上DP, 交了1发, 发现会算重复

贴一下题解的做法

#include <iostream>
#include <algorithm>
#include <cstdio>
#define REP(i,a,n) for(int i=a;i<=n;++i)
using namespace std;

const int N = 510, INF = 0x3f3f3f3f;
int n, m;
struct {int u,v,w;} e[N*N];
int d[N][N], cnt[N], ans[N][N];

int main() {
	scanf("%d%d", &n, &m);
	memset(d, 0x3f, sizeof d);
	REP(i,1,n) d[i][i] = 0;
	REP(i,1,m) {
		int x, y, l;
		scanf("%d%d%d", &x, &y, &l);
		d[x][y] = d[y][x] = l;
		e[i] = {x,y,l};
	}
	REP(k,1,n) REP(i,1,n) REP(j,1,n) {
		d[i][j] = min(d[i][j],d[i][k]+d[k][j]);
	}
	REP(i,1,n) {
		memset(cnt, 0, sizeof cnt);
		REP(j,1,m) {
			if (d[i][e[j].u]+e[j].w==d[i][e[j].v]) ++cnt[e[j].v];
			if (d[i][e[j].v]+e[j].w==d[i][e[j].u]) ++cnt[e[j].u];
		}
		REP(j,1,n) REP(k,1,n) {
			if (d[i][k]+d[k][j]==d[i][j]) {
				ans[i][j]+=cnt[k];
			}
		}
	}
	REP(i,1,n) REP(j,i+1,n) printf("%d ",ans[i][j]);
	puts("");
}