Codeforces Round #372 (Div. 1) B. Complete The Graph （枚举+最短路）

题目就是给你一个图，图中部分边没有赋权值，要求你把无权的边赋值，使得s->t的最短路为l。

卡了几周的题了，最后还是经群主大大指点……做出来的……

思路就是跑最短路，然后改权值为最短路和L的差值，直到最短路为L，或者每条无权边都赋值为止。

点是0～n-1的，因为这个错了好几次= =

dijkstra超时，spfa卡过。

看到很多题解说二分，但是实在不能理解那个思路阿…………

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
const int N = 1005;
const ll INF = 1e13+7;

int n, m, l, s, t;
struct edge { int from, to, next, w; } e[N*N];
int head[N], cntE;
void addedge(int u, int v, int w) {
    e[cntE].from = u; e[cntE].to = v; e[cntE].next = head[u]; e[cntE].w = w; head[u] = cntE++;
    e[cntE].from = v; e[cntE].to = u; e[cntE].next = head[v]; e[cntE].w = w; head[v] = cntE++;
}

int cnt[N], pre[N];
ll d[N];
bool inq[N];
int spfa() {
    queue<int> q;
    memset(inq, 0, sizeof inq);
    memset(cnt, 0, sizeof cnt);
    for (int i = 0; i <= n; ++i) d[i] = INF;
    d[s] = 0;
    inq[s] = true;
    q.push(s);

    while (q.size()) {
        int u = q.front(); q.pop();
        inq[u] = false;
        for (int i = head[u]; ~i; i = e[i].next) {
            int v = e[i].to;
            int c = e[i].w;
            if (d[u] < INF && d[v] > d[u] + c) {
                d[v] = d[u] + c;
                pre[v] = u;
                if (!inq[v]) {
                    q.push(v); inq[v] = true;
                }
            }
        }
    }
    return d[t];
}

vector<int> zeroedge;
int main()
{
    //freopen("in.txt", "r", stdin);
    while (~scanf("%d%d%d%d%d", &n, &m, &l, &s, &t)) {
        memset(head, -1, sizeof head); cntE = 0; zeroedge.clear();
        int u, v, c;
        for (int i = 0; i < m; ++i) {
            scanf("%d%d%d", &u, &v, &c);
            addedge(u, v, c==0?1:c);
            if (c == 0) zeroedge.push_back(cntE-1);
        }
        ll tmp = spfa();
        if (tmp > l) {
            puts("NO");
            continue;
        }
        if (tmp == l) { // may be dont have zero edge
            puts("YES");
            for (int i = 0; i < cntE; i += 2) {
                printf("%d %d %d\n", e[i].from, e[i].to, e[i].w);
            }
            continue;
        }
        int fg = false;
        for (int i = 0; i < zeroedge.size(); ++i) {
            int x = zeroedge[i];
            e[x].w = e[x ^ 1].w = l - tmp + 1;
            tmp = spfa();
            if (tmp == l) {
                fg = true;
                puts("YES");
                for (int i = 0; i < cntE; i += 2) {
                    printf("%d %d %d\n", e[i].from, e[i].to, e[i].w);
                }
                break;
            }

        }
        if (!fg) puts("NO");
    }
    return 0;
}