POJ

POJ - 3169 题目链接

思路

知道差分约束的都知道，这是一道线性差分约束裸题
对于ml的数据我们有，dis[y] - dis[x] <= w，于是这里建立边 x -> y，value = w
对于md的数据我们有，dis[y] - dis[x] >= w，变形dis[x] - dis[y] <= -w，建立边 y -> x，value = -1
由于题目有限制条件，两头牛可以放在一起，并且编号小的放在前面，有 dis[i + 1] - dis[i] >= 0，变形dis[i] - dis[i + 1] <= 0，建立边 i + 1 -> i，value = 0

代码

Powered by CK 2020:04:07
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 10;
int head[N], to[N], nex[N], value[N], cnt;
int visit[N], dis[N], num[N], n, ml, md;
void add(int x, int y, int w) {
    to[cnt] = y;
    value[cnt] = w;
    nex[cnt] = head[x];
    head[x] = cnt++;
}
int spfa() {
    memset(dis, 0x3f, sizeof dis);
    memset(visit, 0, sizeof visit);
    memset(num, 0, sizeof num);
    queue<int> q;
    q.push(1);
    dis[1] = 0;
    visit[1] = 1, num[1] = 1;
    while(!q.empty()) {
        int temp = q.front();
        q.pop();
        visit[temp] = 0;
        for(int i = head[temp]; i; i = nex[i]) {
            if(dis[to[i]] > dis[temp] + value[i]) {
                dis[to[i]] = dis[temp] + value[i];
                if(++num[to[i]] == n)   return -1;
                q.push(to[i]);
                visit[to[i]] = 1;
            }
        }
    }
    return dis[n] == INF ? -2 : dis[n];
}
int main() {
    // freopen("in.txt", "r", stdin);
    int x, y, w;
    while(scanf("%d %d %d", &n, &ml, &md) != EOF) {
        memset(head, 0, sizeof head);
        cnt = 1;
        for(int i = 0; i < ml; i++) {
            scanf("%d %d %d", &x, &y, &w);
            add(x, y, w);
        }
        for(int i = 0; i < md; i++) {
            scanf("%d %d %d", &x, &y, &w);
            add(y, x, -w);
        }
        for(int i = 1; i < n; i++)
            add(i + 1, i, 0);
        printf("%d
", spfa());
    }
    return 0;
}