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本来在想 单源多点很好解决但是多源单点怎么解 然后我发现只要倒过来就可以了
把输入存下来然后 处理完dis1 重新init一次 倒着再输入一次 处理dis2 输出max(dis1[i] + dis2[i])
#include <iostream> #include <string> #include <cstdio> #include <cmath> #include <cstring> #include <queue> #include <map> #include <vector> #include <set> #include <algorithm> #define INF 0x3F3F3F3F using namespace std; typedef pair<int, int> pii; struct cmp{ bool operator ()(const pii a, const pii b){ return a.first > b.first; } }; int size, head[1010], point[100010], next[100010], val[100010]; void init() { size = 0; memset(head, -1, sizeof head); } void add(int from, int to, int value) { point[size] = to; val[size] = value; next[size] = head[from]; head[from] = size++; } void dij(int dis[], int s) { priority_queue <pii, vector<pii>, cmp> q; q.push(make_pair(0, s)); dis[s] = 0; while(!q.empty()){ pii u = q.top(); q.pop(); if(dis[u.second] < u.first) continue; for(int i = head[u.second]; ~i; i = next[i]){ int j = point[i]; if(dis[j] > val[i] + u.first){ dis[j] = val[i] + u.first; q.push(make_pair(dis[j], j)); } } } } int main() { int n, m, x;//parameter int site1[100010], site2[100010], value[100010];//data int dis1[1010], dis2[1010], ans;//result //work out dis1 scanf("%d%d%d", &n, &m, &x); memset(dis1, 0x3f, sizeof dis1); memset(dis2, 0x3f, sizeof dis2); init(); for(int i = 0; i < m; i++){ scanf("%d%d%d", &site1[i], &site2[i], &value[i]); add(site1[i], site2[i], value[i]); } dij(dis1, x); //work out dis2 init(); for(int i = 0; i < m; i++){ add(site2[i], site1[i], value[i]); } dij(dis2, x); //output ans = dis1[1] + dis2[1]; for(int i = 2; i <= n; i++){ ans = max(ans, dis1[i] + dis2[i]); //printf("dis1[%d] = %d dis2[%d] = %d\n", i, dis1[i], i, dis2[i]); } printf("%d\n", ans); return 0; }
kuangbin_ShortPathD (POJ 3268)
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原文地址:http://www.cnblogs.com/quasar/p/5077075.html