Invitation Cards POJ - 1511【最短路】

Invitation Cards POJ - 1511

题目

思路：建两张图，一张正向一张反向，先在正向图里求一遍一号点到所有点的距离，再在反向图里求一遍，反向图里一号点到所有点的距离的意思是所有点到1号点的最短距离，最后相加和就是答案

具体代码如下

#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>
#include<vector>
#define x first
#define y secondusing namespace std;typedef long long LL;
typedef pair<LL, LL> PLL;const int N = 1e6 + 10;int n, m;
LL h1[N], e1[N], w1[N], ne1[N], idx1;
LL h2[N], e2[N], w2[N], ne2[N], idx2;
LL dis1[N], dis2[N];
bool st[N];void add1(int a, int b, int c){e1[idx1] = b, w1[idx1] = c, ne1[idx1] = h1[a], h1[a] = idx1++;
}void add2(int a, int b, int c){e2[idx2] = b, w2[idx2] = c, ne2[idx2] = h2[a], h2[a] = idx2++;
}void dijkstra1(){memset(st, 0, sizeof st);memset(dis1, 0x3f, sizeof dis1);dis1[1] = 0;priority_queue<PLL, vector<PLL>, greater<PLL> >heap;heap.push({0, 1});while(heap.size()){PLL t = heap.top();heap.pop();LL ver = t.y;if(st[ver]) continue;st[ver] = true;for(LL i=h1[ver]; i!=-1; i=ne1[i]){LL j = e1[i];if(dis1[j] > dis1[ver] + w1[i]){dis1[j] = dis1[ver] + w1[i];heap.push({dis1[j], j});}}}}void dijkstra2(){memset(st, 0, sizeof st);memset(dis2, 0x3f, sizeof dis2);dis2[1] = 0;priority_queue<PLL, vector<PLL>, greater<PLL> >heap;heap.push({0, 1});while(heap.size()){PLL t = heap.top();heap.pop();LL ver = t.y;if(st[ver]) continue;st[ver] = true;for(LL i=h2[ver]; i!=-1; i=ne2[i]){LL j = e2[i];if(dis2[j] > dis2[ver] + w2[i]){dis2[j] = dis2[ver] + w2[i];heap.push({dis2[j], j});}}}}int main(){int T;scanf("%d", &T);while(T--){memset(h1, -1, sizeof h1);memset(h2, -1, sizeof h2);scanf("%d%d", &n, &m);while(m--){int a, b, c;scanf("%d%d%d", &a, &b, &c);add1(a, b, c);add2(b, a, c);}dijkstra1();dijkstra2();LL res = 0;for(int i=2; i<=n; ++i) res += dis1[i] + dis2[i];printf("%lld\n", res);}return 0;
}

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