UVa 11374

题目链接:https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=0&problem=2369&mosmsg=Submission+received+with+ID+26582716

可以修改一条边的权值,求起点到终点的最短路

从起点和终点分别做一次最短路,枚举修改权值的边是哪一条,新的最短路长度就是两边的最短路加上新的权值

还要输出路径,更新时记录前驱即可

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;

const int maxn = 10010; 
const int INF = 0x3f3f3f3f;

int n, s, t, m, k;
int u[maxn], v[maxn], w[maxn];

int h[maxn], cnt = 0;
struct E{
	int to, cost, next;
}e[maxn];
void add(int u, int v, int w){
	e[++cnt].to = v;
	e[cnt].cost = w;
	e[cnt].next = h[u];
	h[u] = cnt;
}

vector<int> path;
int d[2][maxn];
int pre[2][maxn];
void dij(int S, int type){
	priority_queue<pii, vector<pii>, greater<pii> > q;
	d[type][S] = 0;
	q.push(pii(0, S));
	
	while(!q.empty()){
		pii p = q.top(); q.pop();
		int u = p.second;
		
		if(p.first != d[type][u]) continue; 
		
		for(int i = h[u] ; i != -1 ; i = e[i].next){
			int v = e[i].to;
			if(d[type][v] > d[type][u] + e[i].cost){
				d[type][v] = d[type][u] + e[i].cost;
				pre[type][v] = u;
				q.push(pii(d[type][v], v));
			}
		}
	}
}

ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }

int main(){
	int flag = 0;
	while(scanf("%d%d%d", &n, &s, &t) == 3){
		if(flag) printf("
");
		flag = 1;
		path.clear();
		memset(pre, 0, sizeof(pre));
		memset(h, -1, sizeof(h)); cnt = 0;
		
		scanf("%d", &m);
		for(int i = 1 ; i <= m ; ++i){
			int u, v, w;
			scanf("%d%d%d", &u, &v, &w);
			add(u, v, w); add(v, u, w);
		}
		
		scanf("%d", &k);
		for(int i = 1 ; i <= k ; ++i){
			scanf("%d%d%d", &u[i], &v[i], &w[i]);
		} 
		
		memset(d, 0x3f, sizeof(d));
		dij(s, 0);
		dij(t, 1);
		
		int ans = d[0][t];
		int sta = -1, stb = -1;
		for(int i = 1 ; i <= k ; ++i){
			if(d[0][u[i]] + d[1][v[i]] + w[i] < ans){
				ans = d[0][u[i]] + d[1][v[i]] + w[i];
				sta = u[i]; stb = v[i];
			}
			if(d[0][v[i]] + d[1][u[i]] + w[i] < ans){
				ans = d[0][v[i]] + d[1][u[i]] + w[i];
				sta = v[i]; stb = u[i];
			}
		}
		
		if(sta != -1){
			int cur = sta;
			while(1){
				path.push_back(cur);
				if(cur == s) break;
				cur = pre[0][cur];
			}
			for(int i = path.size() - 1 ; i >= 0 ; --i){
				printf("%d ", path[i]);
			} 
			path.clear();
	
			cur = stb;
			while(1){
				path.push_back(cur);
				if(cur == t) break;
				cur = pre[1][cur];
			}
			for(int i = 0 ; i <= path.size() - 1 ; ++i){
				if(i > 0) printf(" ");
				printf("%d", path[i]);
			} printf("
");
		} else{
			int cur = t;
			while(1){
				path.push_back(cur);
				if(cur == s) break;
				cur = pre[0][cur];
			}
			
			for(int i = path.size() - 1 ; i >= 0 ; --i){
				if(i < path.size()-1) printf(" ");
				printf("%d", path[i]);
			} printf("
");
		}
		if(sta != -1) printf("%d
", sta);
		else printf("Ticket Not Used
");
		printf("%d
", ans);
	}
	return 0;
}
原文地址:https://www.cnblogs.com/tuchen/p/15025570.html