Codeforces 716D(最短路+枚举加边)

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Codeforces 716D
题意：$n$个点$0~n-1$和$m$条边的无向图，每条边上有一个正权或0，0可以改成任何正权，求一个方案使得$s$到$t$最短路为$L$.
解：考虑将 0 边先全部删除，然后再进行添加。
先对原图进行一次最短路，如果此时$s$到$t$最短路小于$L$，显然无解，如果等于则直接输出
否则我们开始随意加边，加最小正边权为$1$，注意正边权。然后每次加完边以后做最短路，如果当前最短路小于$L$，则可以直接输出了，但是这条边要改成适当的权来使最短路变为$L$。

1、枚举最短路要考虑2/4种情况
2、要挖掘题目中所有的限制
3、要严格按照做题步骤去想，先写题意，再想题目类型，五大思维方式

#include<cstdio> 
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<climits>
#include<queue>
#include<cmath>
#define ms(i, j) memset(i, j, sizeof i)
#define LL long long
#define db double
#define fir first
#define sec second
#define mp make_pair
using namespace std;

namespace flyinthesky {

    const int MAXN = 1000 + 5, MAXM = 10000 + 5;

    struct edge {int u, v, w, nxt;} ed[MAXM * 2];
    struct zoe {int s, t;} whw[MAXM]; 
    struct data {
        int u；
        LL dis;
        bool operator < (const data &b) const {return dis > b.dis;}
    };

    priority_queue<data > q;
    int n, m, L, s, t, en, orz, hd[MAXN], cnt;
    int vis[MAXN];
    LL dis[MAXN];

    void ins(int x, int y, int w) {ed[++en] = (edge){x, y, w, hd[x]}, hd[x] = en;}

    void dij(int ss) {
        for (int i = 0; i <= n; ++i) vis[i] = 0, dis[i] = LLONG_MAX / 2;
        dis[ss] = 0, q.push((data){ss, 0});
        while (!q.empty()) {
            data p = q.top(); q.pop();
            if (vis[p.u]) continue;
            vis[p.u] = true;
            for (int i = hd[p.u]; i > 0; i = ed[i].nxt) {
                edge &e = ed[i];
                if (dis[e.v] > dis[p.u] + e.w) {
                    dis[e.v] = dis[p.u] + e.w;
                    q.push((data){e.v, dis[e.v]});
                }
            }
        }
    }
    void pri(int hh) {
        printf("YES\n");
        for (int i = 1; i <= orz; i += 2) printf("%d %d %d\n", ed[i].u, ed[i].v, ed[i].w);
        for (int i = 1; i < hh; ++i) printf("%d %d %d\n", whw[i].s, whw[i].t, 1);
        dij(s);
        LL tmp1 = dis[whw[hh].s], tmp2 = dis[whw[hh].t];
        dij(t);
        tmp1 += dis[whw[hh].t], tmp2 += dis[whw[hh].s];
        if (hh >= 1) printf("%d %d %lld\n", whw[hh].s, whw[hh].t, L - min(tmp1, tmp2));
        for (int i = hh + 1; i <= cnt; ++i) printf("%d %d %lld\n", whw[i].s, whw[i].t, 10000000000ll);
    }

    void clean() {
        orz = cnt = en = 0, ms(hd, -1);
    }
    int solve() {
        cin >> n >> m >> L >> s >> t;
        clean();
        for (int x, y, w, i = 1; i <= m; ++i) {
            scanf("%d%d%d", &x, &y, &w);
            if (w != 0) ins(x, y, w), ins(y, x, w);
            else whw[++cnt] = (zoe){x, y};
        }
        orz = en;
        dij(s);
        if (dis[t] < L) return printf("NO\n"), 0;
        if (dis[t] == L) return pri(0), 0;
        int gg = 0;
        for (int i = 1; i <= cnt; ++i) {
            ins(whw[i].s, whw[i].t, 1), ins(whw[i].t, whw[i].s, 1);
            dij(s);
            if (dis[t] <= L) {gg = i; break ;}
        }
        if (dis[t] <= L) pri(gg); else printf("NO\n");
        return 0; 
    }
}
int main() {
    flyinthesky::solve();
    return 0;
}
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