kuangbin_ShortPath D (POJ 3268)

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;
}