题目描述
给定一个n个点m条边的有向图,图中可能存在重边和自环,所有边权均为非负值。
请你求出1号点到n号点的最短距离,如果无法从1号点走到n号点,则输出-1。
输入格式
第一行包含整数n和m。
接下来m行每行包含三个整数x,y,z,表示存在一条从点x到点y的有向边,边长为z。
输出格式
输出一个整数,表示1号点到n号点的最短距离。
如果路径不存在,则输出-1。
数据范围
1≤n,m≤1.5×105,
图中涉及边长均不小于0,且不超过10000。
样例
输入样例:
3 3
1 2 2
2 3 1
1 3 4
输出样例:
3
算法1(超时)
(朴素dijkstra算法) $O(n^2)$
C++ 代码
#include <stdio.h>
#include <string.h>
#include <math.h>
const int N = 1e6, M = 2e6;
int dist[N];
int h[N], to[M], w[M], ne[M], idx;
bool st[N];
int n, m;
void add(int a, int b, int c){
to[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}
void dijkstra(int s){
for(int i = h[s]; ~i; i = ne[i]) dist[to[i]] = fmin(dist[to[i]], w[i]);
dist[s] = 0;
st[s] = true;
while(true){
int mind = 0x3f3f3f3f, v = -1;
for(int i = 1; i <= n; i++){
if(!st[i] && dist[i] < mind){
mind = dist[i];
v = i;
}
}
if(v == -1) break;
st[v] = true;
for(int i = h[v]; ~i; i = ne[i]){
dist[to[i]] = fmin(dist[to[i]], dist[v] + w[i]);
}
}
}
int main(void){
int a, b, c;
scanf("%d %d", &n, &m);
memset(dist, 0x3f, sizeof(dist));
memset(h, -1, sizeof(h));
for(int i = 0; i < m; i++){
scanf("%d %d %d", &a, &b, &c);
add(a, b, c);
}
dijkstra(1);
if(dist[n] == 0x3f3f3f3f) puts("-1");
else printf("%d\n", dist[n]);
return 0;
}
算法2(通过)
(堆优化的dijkstra) $O(mlogn)$
C++ 代码
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <limits.h>
const int N = 1e6, M = 2e6;
int dist[N];
int h[N], to[M], w[M], ne[M], idx;
bool st[N];
int heap[M][2];
int re[2]; //分别存从堆中取的点和距离
int n, m, hsize;
void add(int a, int b, int c){
to[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}
void push(int v, int d){
int i;
for(i = ++hsize; heap[i / 2][1] > d; i /= 2){
heap[i][0] = heap[i / 2][0];
heap[i][1] = heap[i / 2][1];
}
heap[i][0] = v;
heap[i][1] = d;
}
void pop(){
re[0] = heap[1][0];
re[1] = heap[1][1];
int lastv = heap[hsize][0];
int lastd = heap[hsize--][1];
int i, child;
for(i = 1; i * 2 <= hsize; i = child){
child = i * 2;
if(child != hsize && heap[child + 1][1] < heap[child][1]) child++;
if(lastd > heap[child][1]){
heap[i][0] = heap[child][0];
heap[i][1] = heap[child][1];
}
else break;
}
heap[i][0] = lastv;
heap[i][1] = lastd;
}
void dijkstra(int s){
dist[s] = 0;
push(s, 0);
while(hsize > 0){
pop();
int v = re[0], dis = re[1];
if(st[v]) continue;
st[v] = true;
for(int i = h[v]; ~i; i = ne[i]){
if(!st[to[i]] && dis + w[i] < dist[to[i]]){
dist[to[i]] = dis + w[i];
push(to[i], dist[to[i]]);
}
}
}
}
int main(void){
int a, b, c;
scanf("%d %d", &n, &m);
h[0] = INT_MIN;
memset(dist, 0x3f, sizeof(dist));
memset(h, -1, sizeof(h));
for(int i = 0; i < m; i++){
scanf("%d %d %d", &a, &b, &c);
add(a, b, c);
}
dijkstra(1);
if(dist[n] == 0x3f3f3f3f) puts("-1");
else printf("%d\n", dist[n]);
return 0;
}