洛谷 [P1608] 最短路计数

最短路计数模版

本题要注意重边的处理

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <cstring>
using namespace std;
const int MAXN = 2005;
int head[MAXN], n, m, nume, dist[MAXN], cnt[MAXN];
bool f[MAXN];
struct edge{
	int to, nxt, dis;
}e[MAXN * MAXN];
void adde(int from, int to, int dis) {
	e[++nume].to = to;
	e[nume].nxt = head[from];
	e[nume].dis = dis;
	head[from] = nume;
}
int init() {
	int rv = 0, fh = 1;
	char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') fh = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		rv = (rv<<1) + (rv<<3) + c - '0';
		c = getchar();
	}
	return fh * rv;
}
struct node{
	int num, dis;
	bool operator < (const node & b) const{
		return dis > b.dis;
	}
}a[MAXN];
priority_queue <node> q;
void dij() {
	memset(dist, 0x3f, sizeof(dist));
	dist[1] = 0;
	q.push((node){1, 0});
	while(!q.empty()) {
		node u = q.top(); q.pop();
		if(f[u.num]) continue;
		f[u.num] = 1;
		for(int i = head[u.num]; i; i = e[i].nxt) {
			node v;
			v.num = e[i].to;
			if(dist[v.num] > dist[u.num] + e[i].dis) {
				dist[v.num] = dist[u.num] + e[i].dis;
				v.dis = dist[v.num];
				q.push(v);
			} 
		}
	}
}
void cnnt() {
	memset(f, 0, sizeof(f));
	for(int i = 1; i <= n; i++) a[i] = (node) {i, dist[i]};
	sort(a + 1, a + n + 1);
	cnt[1] = 1;
	for(int i = n; i >= 1; i--) {
		int u = a[i].num;
		for(int i = head[u]; i; i = e[i].nxt) {
			int v = e[i].to;
			if(dist[v] == dist[u] + e[i].dis) {
				cnt[v] += cnt[u];
			}
		}
	}
}
int ddd[MAXN][MAXN];
int main() {
	n = init(); m = init();
	memset(ddd, 0x3f, sizeof(ddd));
	for(int i = 1; i <= m; i++) {
		int u = init(), v = init(), dis = init();
		ddd[u][v] = min(ddd[u][v], dis);
	}
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= n; j++) {
			if(ddd[i][j] != 0x3f3f3f3f) adde(i, j, ddd[i][j]);
		}
	}
	dij();
	cnnt();
	if(dist[n] == 0x3f3f3f3f) printf("No answer
");
	else printf("%d %d
", dist[n], cnt[n]);
	return 0;
}