poj3159 Candies(差分约束，dij+heap)

poj3159 Candies

这题实质为裸的差分约束。

先看最短路模型：若d[v] >= d[u] + w, 则连边u->v，之后就变成了d[v] <= d[u] + w , 即d[v] – d[u] <= w。

再看题目给出的关系：b比a多的糖果数目不超过c个，即d[b] – d[a] <= c ，正好与上面模型一样，

所以连边a->b，最后用dij+heap求最短路就行啦。

ps:我用vector一直TLE，后来改用链式前向星才过了orz。。。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#define CLR(a,b) memset((a),(b),sizeof((a)))
using namespace std;

const int N = 30001;
const int M = 150001;
const int inf = 0x3f3f3f3f;

int n, m;
bool s[N];
int head[N];
int cnt;
struct edge{
    int nex;
    int v, w;
}g[M];
struct node{
    int v, w;
    node(int _v=0,int _w=0):v(_v),w(_w){}
    bool operator < (const node&r)const{
        return r.w < w;
    }
};
void add_edge(int u,int v,int w){
    g[cnt].v = v;
    g[cnt].w = w;
    g[cnt].nex = head[u];
    head[u] = cnt++;
}
void dij(){
    int i, u, v, w;
    node t;
    CLR(s, 0);
    priority_queue<node>q;
    q.push(node(1,0));
    while(!q.empty()){
        t = q.top(); q.pop();
        u = t.v;
        if(s[u]) continue;
        s[u] = 1;
        if(u == n) break;
        for(i = head[u]; ~i; i = g[i].nex){
            v = g[i].v;
            if(!s[v]){
                w = t.w + g[i].w;
                q.push(node(v, w));
            }
        }
    }
    printf("%d\n", t.w);
}
int main(){
    int i, j, a, b, c;
    scanf("%d %d", &n, &m);
    cnt = 0;
    CLR(head, -1);
    for(i = 1; i <= m; ++i){
        scanf("%d %d %d", &a, &b, &c);
        add_edge(a,b,c);
    }
    dij();
    return 0;
}